Evaluate the integral. ∫ − 6 x 2 + 3 x + 5 x 3 − x d x \int_{}^{}\frac{-6x^2+3x+5}{x^3-x}dx

Evaluate the integral. ∫ 4 x 2 + 2 x + 3 x 2 + x d x \int_{}^{}\frac{4x^2+2x+3}{x^2+x}dx

4 x + 3 ln ⁡ ∣ x ∣ − 5 ln ⁡ ∣ x + 1 ∣ + C 4x+3\ln\left|x\right|-5\ln\left|x+1\right|+C

4 x + 2 ln ⁡ ∣ x ∣ + 3 ln ⁡ ∣ x − 1 ∣ + C 4x+2\ln\left|x\right|+3\ln\left|x-1\right|+C

4 x + 3 ln ⁡ ∣ x ∣ + 5 ln ⁡ ∣ x + 1 ∣ + C 4x+3\ln\left|x\right|+5\ln\left|x+1\right|+C

4 x + 2 ln ⁡ ∣ x ∣ − 3 ln ⁡ ∣ x + 1 ∣ + C 4x+2\ln\left|x\right|-3\ln\left|x+1\right|+C

Use the method of partial fractions to evaluate the integral. ∫ 5 x 3 − x 2 + 7 x − 2 ( x 2 + 1 ) ( x 2 + 2 ) d x \int_{}^{}\frac{5x^3-x^2+7x-2}{(x^2+1)(x^2+2)}dx

ln ⁡ ∣ x 2 + 1 ∣ − t a n − 1 ( x ) + 3 2 ln ⁡ ∣ x 2 + 2 ∣ + C \ln\left|x^2+1\right|-tan^{-1}\left(x\right)+\frac32\ln\left|x^2+2\right|+C

5 ln ⁡ ∣ x 2 + 1 ∣ − t a n − 1 ( x ) + 3 ln ⁡ ∣ x 2 + 2 ∣ + C 5\ln\left|x^2+1\right|-tan^{-1}\left(x\right)+3\ln\left|x^2+2\right|+C

Evaluate the integral. ∫ t 4 + t 2 − 2 t 3 + t d t \int_{}^{}\frac{t^4+t^2-2}{t^3+t}dt

t 2 2 − 2 ln ⁡ ∣ t ∣ + ln ⁡ ∣ t 2 + 1 ∣ + C \frac{t^2}{2}-2\ln\left|t\right|+\ln\left|t^2+1\right|+C

t 2 2 − 2 ln ⁡ ∣ t ∣ + t a n − 1 ( t ) + C \frac{t^2}{2}-2\ln\left|t\right|+tan^{-1}\left(t\right)+C

t 2 2 − 2 ln ⁡ ∣ t ∣ − t a n − 1 ( t ) + C \frac{t^2}{2}-2\ln\left|t\right|-tan^{-1}\left(t\right)+C

t 2 2 − 2 ln ⁡ ∣ t ∣ − ln ⁡ ∣ t 2 + 1 ∣ + C \frac{t^2}{2}-2\ln\left|t\right|-\ln\left|t^2+1\right|+C

12. Techniques of Integration

Partial Fractions

12. Techniques of Integration

Partial Fractions: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions, which is essential for integration. When dealing with distinct linear factors, the function can be decomposed into constants over each factor. For repeated linear factors, include terms for each power. In cases of irreducible quadratic factors, the numerators become linear expressions. The process involves strategic substitutions or systems of equations to find constants, facilitating easier integration of complex rational functions.

concept

Partial Fraction Decomposition: Distinct Linear Factors

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10m

Partial Fraction Decomposition: Distinct Linear Factors Video Summary

Partial fraction decomposition is a technique used to break down a single rational function into a sum of simpler fractions. This method is particularly useful for integrating rational functions in calculus. The process begins with a rational function expressed as $ \frac{p(x)}{q(x)} $, where $ p(x) $ is the numerator and $ q(x) $ is the denominator. The goal is to express this function as a sum of fractions with distinct linear factors in the denominator.

To start, factor the denominator $ q(x) $ into distinct linear factors. For example, if $ q(x) $ factors into $ (x + 3)(x + 5) $, the partial fraction decomposition can be set up as:

\[\frac{p(x)}{(x + 3)(x + 5)} = \frac{A}{x + 3} + \frac{B}{x + 5}\]

Here, $ A $ and $ B $ are constants that need to be determined. To find these constants, multiply both sides of the equation by the common denominator $ (x + 3)(x + 5) $ to eliminate the fractions:

\[p(x) = A(x + 5) + B(x + 3)\]

Next, expand the right side and group like terms. This will yield an equation that can be compared to the original polynomial $ p(x) $. By equating the coefficients of corresponding powers of $ x $, you can create a system of equations. For instance, if $ p(x) = 2x + 14 $, you would set up the following equations based on the coefficients:

\[A + B = 2\]

\[5A + 3B = 14\]

Solving this system can be done using substitution or elimination methods. For example, if you multiply the first equation by 3 and subtract it from the second, you can isolate one variable and solve for it. Once you find the values of $ A $ and $ B $, substitute them back into the partial fractions to complete the decomposition.

In cases where the degree of the numerator $ p(x) $ is greater than or equal to the degree of the denominator $ q(x) $, the rational function is considered improper. In such instances, perform polynomial long division first to rewrite the function in a proper form before applying partial fraction decomposition.

This method of partial fraction decomposition is essential for simplifying complex rational expressions, making it easier to integrate or analyze them in various mathematical contexts.

example

Partial Fraction Decomposition: Distinct Linear Factors Example 1

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Partial Fraction Decomposition: Distinct Linear Factors Example 1 Video Summary

To find the partial fraction decomposition of the rational function $\frac{3x^2 + 3x - 11}{x^2 + 3x - 10}$, we first need to check the degrees of the numerator and denominator. Since both have a degree of 2, we must perform polynomial division before proceeding with the decomposition.

We start by dividing $3x^2 + 3x - 11$ by $x^2 + 3x - 10$. The first step in polynomial division is to determine what to multiply $x^2$ by to obtain $3x^2$, which is 3. We then multiply the entire denominator by 3:

$3(x^2 + 3x - 10) = 3x^2 + 9x - 30$.

Next, we subtract this result from the numerator:

$3x^2 + 3x - 11 - (3x^2 + 9x - 30) = -6x + 19$.

This gives us the expression:

$\frac{3x^2 + 3x - 11}{x^2 + 3x - 10} = 3 + \frac{-6x + 19}{x^2 + 3x - 10}$.

Now, we need to factor the denominator $x^2 + 3x - 10$. We look for two numbers that multiply to $-10$ and add to $3$, which are $5$ and $-2$. Thus, we can rewrite the denominator as:

$x^2 + 3x - 10 = (x + 5)(x - 2)$.

Now we can express the fraction as:

$\frac{-6x + 19}{(x + 5)(x - 2)}$.

Next, we set up the partial fraction decomposition:

$\frac{-6x + 19}{(x + 5)(x - 2)} = \frac{A}{x + 5} + \frac{B}{x - 2}$.

To find constants $A$ and $B$, we multiply both sides by the common denominator $(x + 5)(x - 2)$ to eliminate the fractions:

$-6x + 19 = A(x - 2) + B(x + 5)$.

Expanding the right side gives:

\(-6x + 19 = Ax - 2A + Bx + 5B = (A + B)x + (-2A + 5B).\

Now, we can equate coefficients from both sides:

1. $A + B = -6$ (coefficient of $x$)

2. $-2A + 5B = 19$ (constant term)

We can solve this system of equations. From the first equation, we can express $A$ in terms of $B$:

A = -6 - B.

Substituting this into the second equation gives:

$-2(-6 - B) + 5B = 19$

12 + 2B + 5B = 19

7B = 7

B = 1.

Now substituting $B = 1$ back into the first equation:

A + 1 = -6

A = -7.

Now that we have $A$ and $B$, we can rewrite the partial fraction decomposition:

$\frac{-6x + 19}{(x + 5)(x - 2)} = \frac{-7}{x + 5} + \frac{1}{x - 2}$.

Finally, combining everything, the complete expression for the original rational function is:

$3 - \frac{7}{x + 5} + \frac{1}{x - 2}$.

This process illustrates how to perform polynomial division followed by partial fraction decomposition, allowing us to break down complex rational functions into simpler components.

Problem

Express the rational function as a sum or difference of two simpler fractions. Use a system of equations.
$\frac{1}{(2 x + 1) (2 x - 1)}$

$- \frac{1}{2 (2 x + 1)} + \frac{1}{2 (2 x - 1)}$

$\frac{1}{2 x + 1} + \frac{1}{2 x - 1}$

$\frac{1}{2 (2 x + 1)} + \frac{1}{2 (2 x - 1)}$

$\frac{1}{2 x + 1} - \frac{1}{2 x - 1}$

Problem

Express the rational function as a sum or difference of two simpler fractions. Use a system of equations.
$\frac{4 x^{2} - 21 x - 40}{x^{3} - x^{2} - 20 x}$

$\frac{2}{x} - \frac{1}{x - 5} - \frac{3}{x + 4}$

$\frac{1}{x} - \frac{2}{x^{2} - x - 20}$

$\frac{2}{x} + \frac{1}{x + 5} - \frac{3}{x - 4}$

$\frac{2}{x} - \frac{1}{x - 5} + \frac{3}{x + 4}$

concept

Partial Fraction Decomposition: Distinct Linear Factors with Shortcut

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Partial Fraction Decomposition: Distinct Linear Factors with Shortcut Video Summary

In the study of rational functions, partial fraction decomposition is a technique used to express a complex rational function as a sum of simpler fractions. This method can often be tedious, involving multiple steps such as expanding, grouping terms, and solving systems of equations. However, there is a more efficient approach known as the "cover-up method" or "Heaviside method" that simplifies the process significantly.

To begin, consider a rational function f(x) with a factored denominator q(x). The first step is to factor the denominator into distinct linear factors. For example, if the denominator factors into (x + 3)(x + 5), the next step is to set up the partial fractions as:

$$ \frac{A}{x + 3} + \frac{B}{x + 5} $$

where A and B are constants to be determined. Instead of expanding and solving a system of equations, we can strategically substitute values for x that will simplify the calculations. For instance, if we substitute x = -5, the term involving B becomes zero:

$$ f(-5) = A(-5 + 3) + B(-5 + 5) $$

This simplifies to:

$$ f(-5) = A(-2) + 0 $$

From this, we can solve for A directly. Similarly, to find B, we can substitute x = -3, which will eliminate the term involving A:

$$ f(-3) = A(-3 + 3) + B(-3 + 5) $$

This simplifies to:

$$ f(-3) = 0 + B(2) $$

By solving these equations, we can quickly find the values of A and B without the need for extensive algebraic manipulation.

In summary, the cover-up method allows for a more straightforward approach to partial fraction decomposition by using strategic substitutions to isolate constants. This method not only saves time but also reduces the complexity of the calculations involved in finding the partial fractions of a rational function.

Problem

Give the partial fraction decomposition for the following expression using strategic substitutions for $x$ .
$\frac{4 x^{2} - 8 x - 8}{x (x - 2) (x + 2)}$

$\frac{2}{x} - \frac{1}{x - 2} - \frac{3}{x + 2}$

$\frac{2}{x} - \frac{1}{x - 2} + \frac{3}{x + 2}$

$\frac{4}{x^{}} - \frac{8}{x - 2} - \frac{8}{x + 2}$

$\frac{4}{x^{}} - \frac{8}{x - 2} + \frac{8}{x + 2}$

Problem

Express the rational function as a sum or difference or simpler rational expressions.
$\frac{4}{x^{3} - 3 x^{2} - x + 3}$

$\frac{1}{2 (x - 3)} + \frac{1}{2 (x + 1)} - \frac{1}{x - 1}$

$\frac{4}{x^{3}} - \frac{4}{3 x^{2}} - \frac{4}{x + 3}$

$\frac{4}{x^{3}} - \frac{4}{3 x^{2}} + \frac{4}{x + 3}$

$\frac{1}{2 (x - 3)} - \frac{1}{2 (x + 1)} + \frac{1}{x - 1}$

Problem

Express the rational function as a sum or difference or simpler rational expressions.
$\frac{13 x + 2}{2 x^{2} - x - 1}$

$\frac{13}{x - 1} + \frac{2}{2 - x}$

$\frac{13}{x + 1} - \frac{2}{2 - x}$

$\frac{5}{x - 1} + \frac{3}{2 x + 1}$

$\frac{5}{x + 1} + \frac{3}{2 x - 1}$

concept

Integration Using Partial Fractions

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Integration Using Partial Fractions Video Summary

Partial fraction decomposition is a valuable technique used to simplify the integration of rational functions, particularly when other methods, such as variable substitution or integration by parts, are ineffective. This method involves breaking down a complex rational function into simpler fractions that can be integrated more easily.

To illustrate this process, consider the function f(x) = (2x - 5) / (x² - x). The first step is to factor the denominator, which can be expressed as x(x - 1). This allows us to rewrite the function as a sum of two fractions: A/x + B/(x - 1), where A and B are constants to be determined.

Next, we multiply both sides of the equation by the common denominator, x(x - 1), to eliminate the fractions. This results in the equation:

2x - 5 = A(x - 1) + Bx

To find the constants A and B, we can strategically choose values for x that simplify the equation. Setting x = 1 eliminates the A term, allowing us to solve for B:

2(1) - 5 = B(1) ⟹ -3 = B

Next, setting x = 0 eliminates the B term, allowing us to solve for A:

2(0) - 5 = A(0 - 1) ⟹ -5 = -A ⟹ A = 5

With A = 5 and B = -3, we can rewrite the original function as:

f(x) = 5/x - 3/(x - 1)

Now, we can integrate the function:

∫f(x) dx = ∫(5/x - 3/(x - 1)) dx

This can be separated into two integrals:

∫(5/x) dx - ∫(3/(x - 1)) dx

Using the integral rules, we find:

5 ln|x| - 3 ln|x - 1| + C

Thus, the final result of the integral is:

5 ln|x| - 3 ln|x - 1| + C

In summary, when faced with the integration of a rational function, if standard methods do not yield results, consider using partial fraction decomposition. This approach allows for the simplification of the function into manageable parts, making integration straightforward.

example

Integration Using Partial Fractions Example 2

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Integration Using Partial Fractions Example 2 Video Summary

When tackling integrals involving rational functions, it's essential to identify the appropriate method for solving them rather than seeking the final solution immediately. This process involves analyzing the structure of the integrals and determining the best approach based on the characteristics of the numerator and denominator.

For instance, consider the integral of a constant over a polynomial, such as ∫ (3 / (x² - 2x + 17)) dx. In this case, long division is not applicable since the numerator is a constant. Substitution also proves ineffective because the derivative of the denominator does not correspond to the numerator. Instead, completing the square is the most suitable method. By rewriting the denominator as (x - 1)² + 16, we can recognize that this integral can be solved using the inverse tangent function.

In another example, the integral ∫ (4 / (x² - 2x - 15)) dx also begins with a constant numerator. Here, we can factor the denominator into (x + 3)(x - 5), which allows us to apply partial fraction decomposition. This method is effective when the denominator can be expressed as distinct linear factors.

When faced with an integral like ∫ ((2x - 2) / (x² - 2x - 15)) dx, we notice that the degree of the numerator is less than that of the denominator, eliminating the need for long division. Instead, substitution is a viable option since the derivative of the denominator matches the numerator, simplifying the integral to ∫ (1/u) du, which is straightforward to integrate.

In cases where the degree of the numerator exceeds that of the denominator, such as ∫ (x³ / (x² - 2x - 15)) dx, long division becomes necessary. This method allows us to simplify the integral before potentially applying other techniques like substitution or partial fractions.

Ultimately, the key to solving these integrals lies in recognizing the relationships between the degrees of the polynomials involved and the potential for factoring or substitution. By systematically analyzing each integral, we can determine the most effective strategy for integration, whether it be completing the square, partial fraction decomposition, or substitution.

Problem

Evaluate the integral.
$\int_{}^{} \frac{- 6 x^{2} + 3 x + 5}{x^{3} - x} d x$

$- 6 \ln ∣ x ∣ + 3 \ln ∣ x + 1 ∣ + 5 \ln ∣ x - 1 ∣ + C$

$- 5 \ln ∣ x ∣ + 4 \ln ∣ x + 1 ∣ + 7 \ln ∣ x - 1 ∣ + C$

$- 5 \ln ∣ x ∣ + 4 \ln ∣ x - 1 ∣ + 7 \ln ∣ x - 1 ∣ + C$

$- 6 \ln ∣ x ∣ + 3 \ln ∣ x - 1 ∣ + 5 \ln ∣ x + 1 ∣ + C$

Problem

Evaluate the integral.
$\int_{0}^{3} \frac{3 x + 10}{x^{2} + 9 x + 20} d x$

$6.51$

$negative 13.22$

$negative 4.19$

$1.23$

Problem

Evaluate the integral.
$\int_{}^{} \frac{4 x^{2} + 2 x + 3}{x^{2} + x} d x$

$4 x + 2 \ln ∣ x ∣ + 3 \ln ∣ x - 1 ∣ + C$

$4 x + 3 \ln ∣ x ∣ + 5 \ln ∣ x + 1 ∣ + C$

$4 x + 3 \ln ∣ x ∣ - 5 \ln ∣ x + 1 ∣ + C$

$4 x + 2 \ln ∣ x ∣ - 3 \ln ∣ x + 1 ∣ + C$

Problem

Find the area under the curve of $y = \frac{10}{(x + 2) (x + 3)}$ between $x equals 0$ and $x equals 4$ .

$2.51$

$19.46$

$negative 1.54$

$37.38$

example

Integration Using Partial Fractions Example 3

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Integration Using Partial Fractions Example 3 Video Summary

To solve the indefinite integral of the function involving cosine and sine, we start by recognizing the need for a variable substitution. We let $ u = \sin(x) $, which transforms the integral into a more manageable form. The differential $ du $ is derived from the derivative of sine, giving us $ du = \cos(x) \, dx $. This substitution allows us to rewrite the integral in terms of $ u $, specifically as:

\[\int \frac{\cos(x)}{\sin(x)(\sin(x) - 1)} \, dx = \int \frac{1}{u(u - 1)} \, du\]

Next, we apply partial fraction decomposition to the integrand $ \frac{1}{u(u - 1)} $. We express it as:

\[\frac{1}{u(u - 1)} = \frac{A}{u} + \frac{B}{u - 1}\]

Multiplying through by the common denominator $ u(u - 1) $ leads to the equation:

\[1 = A(u - 1) + Bu\]

To find the constants $ A $ and $ B $, we can strategically substitute values for $ u $. Setting $ u = 1 $ gives:

\[1 = A(1 - 1) + B(1) \implies B = 1\]

Next, substituting $ u = 0 $ yields:

\[1 = A(0 - 1) + B(0) \implies 1 = -A \implies A = -1\]

With $ A $ and $ B $ determined, we rewrite the integral as:

\[\int \left( -\frac{1}{u} + \frac{1}{u - 1} \right) \, du\]

This can be integrated separately, resulting in:

\[-\ln|u| + \ln|u - 1| + C\]

Substituting back $ u = \sin(x) $ gives us the final result:

\[-\ln|\sin(x)| + \ln|\sin(x) - 1| + C\]

Thus, the solution to the integral is:

\[\ln\left|\frac{\sin(x) - 1}{\sin(x)}\right| + C\]

This process illustrates the effective use of variable substitution and partial fraction decomposition in solving integrals involving trigonometric functions.

concept

Partial Fraction Decomposition: Repeated Linear Factors

Video duration:

11m

Partial Fraction Decomposition: Repeated Linear Factors Video Summary

In the study of rational functions, understanding how to decompose them into partial fractions is essential, especially when dealing with repeated linear factors. When a rational function has a denominator that includes a linear factor raised to a power, the decomposition process requires a specific approach. For a linear factor of the form $ ax + b $ raised to the power of $ n $, the decomposition must include a fraction for each power from 1 to $ n $. This means that the decomposition will take the form:

\[\frac{A}{ax + b} + \frac{B}{(ax + b)^2} + \ldots + \frac{C}{(ax + b)^n}\]

Here, $ A, B, \ldots, C $ are constants that need to be determined. This differs from the case of distinct linear factors, where each factor contributes only one term to the decomposition.

To illustrate this process, consider a rational function with a denominator that includes both distinct and repeated linear factors. For example, if the denominator is $ (x + 1)(x - 1)^2 $, the partial fraction decomposition would be set up as:

\[\frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}\]

Next, to find the constants $ A, B, $ and $ C $, one effective method is to multiply both sides of the equation by the common denominator, which eliminates the fractions. This results in a polynomial equation that can be solved for the constants. For instance, multiplying through by $ (x + 1)(x - 1)^2 $ leads to an equation that can be simplified to find the values of $ A, B, $ and $ C $.

Strategic substitutions for $ x $ can simplify the process of solving for these constants. By choosing values for $ x $ that will cancel out certain terms, you can isolate and solve for the constants more easily. For example, setting $ x = 1 $ will eliminate terms involving $ (x - 1) $, while setting $ x = -1 $ will eliminate terms involving $ (x + 1) $.

Once the constants are determined, the final step is to rewrite the original rational function using the found values. This results in a complete partial fraction decomposition, which can be used for further analysis or integration.

In summary, when dealing with repeated linear factors in rational functions, it is crucial to include a term for each power of the factor in the decomposition. The process of finding the constants can be efficiently handled through strategic substitutions, leading to a clearer understanding of the function's behavior.

example

Partial Fraction Decomposition: Repeated Linear Factors Example 4

Video duration:

11m

Partial Fraction Decomposition: Repeated Linear Factors Example 4 Video Summary

To evaluate the integral of the function $\frac{x - 1}{x^2(x + 1)}$ using the method of partial fractions, we first need to decompose the fraction into simpler components. The denominator consists of a repeated linear factor $x^2$ and a distinct linear factor $x + 1$. This leads us to express the function as:

\[\frac{x - 1}{x^2(x + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 1}\]

Here, $A$, $B$, and $C$ are constants that we need to determine. To find these constants, we multiply both sides of the equation by the common denominator $x^2(x + 1)$, which simplifies our equation to:

\[x - 1 = A(x)(x + 1) + B(x + 1) + C(x^2)\]

Next, we can strategically choose values for $x$ to eliminate terms and solve for the constants. Setting $x = -1$ gives us:

\[-2 = C\]

Now, substituting $x = 0$ allows us to solve for $B$:

\[-1 = B\]

To find $A$, we can choose another value, such as $x = 1$:

\[0 = 2A - 1 - 2 \implies 2A = 4 \implies A = 2\]

Now that we have determined the constants, we can rewrite the original function as:

\[\frac{x - 1}{x^2(x + 1)} = \frac{2}{x} - \frac{1}{x^2} - \frac{2}{x + 1}\]

Next, we can integrate each term separately:

\[\int \left( \frac{2}{x} - \frac{1}{x^2} - \frac{2}{x + 1} \right) dx\]

The integrals yield:

\[2 \ln |x| + \frac{1}{x} - 2 \ln |x + 1| + C\]

For the definite integral from 1 to 4, we evaluate:

\[\left[ 2 \ln |x| + \frac{1}{x} - 2 \ln |x + 1| \right]_{1}^{4}\]

Calculating this gives:

\[\left( 2 \ln 4 + \frac{1}{4} - 2 \ln 5 \right) - \left( 2 \ln 1 + 1 - 2 \ln 2 \right)\]

Since $\ln 1 = 0$, this simplifies to:

\[2 \ln 4 + \frac{1}{4} - 2 \ln 5 - 1 + 2 \ln 2\]

Combining the logarithmic terms results in:

\[2 \ln \left( \frac{4}{5} \right) + \frac{1}{4} - 1\]

Finally, evaluating this expression will yield an approximate result of 0.19. This process illustrates how to effectively use partial fraction decomposition to solve definite integrals involving rational functions.

Problem

Express the integrand as a sum of partial fractions and evaluate the integral.
$\int_{}^{} \frac{2}{x^{3} + x^{2} - x - 1} d x$

$\int_{}^{} (- \frac{1}{2 (x + 1)} - \frac{1}{{(x + 1)}^{2}} + \frac{1}{2 (x - 1)}) d x = - \frac{1}{2} \ln ∣ x + 1 ∣ + \ln ∣ (x + 1)^{2} ∣ + \frac{1}{2} \ln ∣ x - 1 ∣ + C$

$\int_{}^{} (- \frac{2}{(x + 1)} - \frac{2}{{(x + 1)}^{2}} + \frac{2}{(x - 1)}) d x = - 2 \ln ∣ x + 1 ∣ + \frac{2}{x + 1} + 2 \ln ∣ x - 1 ∣ + C$

$\int_{}^{} (- \frac{1}{2 (x + 1)} - \frac{1}{{(x + 1)}^{2}} + \frac{1}{2 (x - 1)}) d x = - \frac{1}{2} \ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{1}{2} \ln ∣ x - 1 ∣ + C$

$\int_{}^{} (- \frac{2}{(x + 1)} - \frac{2}{{(x + 1)}^{2}} + \frac{2}{(x - 1)}) d x = - 2 \ln ∣ x + 1 ∣ + 2 \ln ∣ {(x + 1)}^{2} ∣ + 2 \ln ∣ x - 1 ∣ + C$

example

Partial Fraction Decomposition: Repeated Linear Factors Example 5

Video duration:

12m

Partial Fraction Decomposition: Repeated Linear Factors Example 5 Video Summary

To find the volume of a solid formed by revolving a region bounded by a function $ f(x) $ around the x-axis, we utilize the formula for volume derived from integration. The volume $ V $ can be calculated using the integral of the area function $ A(x) $ with respect to $ x $. The area function for a solid of revolution is given by:

\[ A(x) = \pi [f(x)]^2 \]

In this case, the function is defined as:

\[ f(x) = \frac{1}{x \sqrt{(x-1)^2}} \]

Squaring this function results in:

\[ A(x) = \pi \cdot \frac{1}{x^2 (x-1)} \]

To find the volume, we set up the integral from $ x = 2 $ to $ x = 4 $:

\[ V = \int_{2}^{4} A(x) \, dx = \int_{2}^{4} \frac{\pi}{x^2 (x-1)} \, dx \]

Factoring out $ \pi $, we have:

\[ V = \pi \int_{2}^{4} \frac{1}{x^2 (x-1)} \, dx \]

To solve this integral, we apply the method of partial fractions. The expression can be decomposed as:

\[ \frac{1}{x^2 (x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} \]

Multiplying through by the common denominator $ x^2 (x-1) $ and equating coefficients allows us to solve for constants $ A $, $ B $, and $ C $. After substituting strategic values for $ x $ (such as $ x = 1 $ and $ x = 0 $), we find:

\[ C = 1, \quad B = -1, \quad A = -1 \]

Thus, the partial fraction decomposition is:

\[ \frac{1}{x^2 (x-1)} = -\frac{1}{x} - \frac{1}{x^2} + \frac{1}{x-1} \]

Now, we can integrate each term separately:

\[ V = \pi \left( -\ln|x| + \frac{1}{x} + \ln|x-1| \right) \bigg|_{2}^{4} \]

Evaluating this from $ x = 2 $ to $ x = 4 $, we compute:

\[ V = \pi \left( -\ln(4) + \frac{1}{4} + \ln(3) \right) - \left( -\ln(2) + \frac{1}{2} + \ln(1) \right) \]

Since $ \ln(1) = 0 $, this simplifies to:

\[ V = \pi \left( -\ln(4) + \frac{1}{4} + \ln(3) + \ln(2) - \frac{1}{2} \right) \]

After simplifying and calculating, the final volume is approximately:

\[ V \approx 0.16 \]

This process illustrates how to find the volume of a solid of revolution using integration, partial fractions, and careful evaluation of the integral bounds.

concept

Partial Fraction Decomposition: Irreducible Quadratic Factors

Video duration:

13m

Partial Fraction Decomposition: Irreducible Quadratic Factors Video Summary

Partial fraction decomposition is a technique used to break down rational functions into simpler fractions. This process becomes particularly important when dealing with irreducible quadratic factors in the denominator. An irreducible quadratic factor is a polynomial of degree two that cannot be factored into linear factors over the real numbers, such as $x^2 + 1$.

When performing partial fraction decomposition with an irreducible quadratic factor, the numerators of the fractions will not be constants but rather linear expressions. For example, if the denominator includes an irreducible quadratic factor $ax^2 + bx + c$, the corresponding numerator will take the form $Ax + B$, where $A$ and $B$ are constants to be determined. If the irreducible quadratic factor is raised to a power $n$, the decomposition will include terms for each power from one to $n$, such as:

\[\frac{Ax + B}{ax^2 + bx + c} + \frac{Cx + D}{(ax^2 + bx + c)^2} + \ldots + \frac{Kx + L}{(ax^2 + bx + c)^n}\]

To illustrate this, consider the rational function:

\[\frac{4x^2 - 3x + 5}{(x - 2)(x^2 + 1)}\]

Here, $x - 2$ is a distinct linear factor, while $x^2 + 1$ is an irreducible quadratic factor. The partial fraction decomposition will be set up as:

\[\frac{A}{x - 2} + \frac{Bx + C}{x^2 + 1}\]

To find the constants $A$, $B$, and $C$, multiply both sides by the common denominator $(x - 2)(x^2 + 1)$ to eliminate the fractions. This results in:

\[4x^2 - 3x + 5 = A(x^2 + 1) + (Bx + C)(x - 2)\]

Next, substitute strategic values for $x$ to simplify the equations. For instance, setting $x = 2$ will eliminate the term involving $A$, allowing you to solve for $A$. After finding $A$, expand the equation and group like terms to create a system of equations based on the coefficients of $x^2$, $x$, and the constant terms. This will yield a set of equations that can be solved to find $B$ and $C$.

Once all constants are determined, substitute them back into the partial fraction decomposition to obtain the final result. This method not only simplifies the rational function but also makes it easier to integrate or analyze further.

In summary, when dealing with irreducible quadratic factors in partial fraction decomposition, remember to use linear numerators and follow a systematic approach to find the constants involved. This technique is essential for simplifying complex rational expressions in calculus and algebra.

example

Partial Fraction Decomposition: Irreducible Quadratic Factors Example 6

Video duration:

16m

Partial Fraction Decomposition: Irreducible Quadratic Factors Example 6 Video Summary

To evaluate the definite integral from 1 to 2 of the function $\frac{x - 5}{(x + 1)(x^2 + 3x - 1)}$ with respect to $x$, we first recognize that we are dealing with a rational function. A variable substitution may not simplify the integral effectively, so we will use partial fraction decomposition instead.

We start by expressing the integrand in terms of simpler fractions. The denominator consists of a linear factor $x + 1$ and an irreducible quadratic factor $x^2 + 3x - 1$. Thus, we can write:

\[\frac{x - 5}{(x + 1)(x^2 + 3x - 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 3x - 1}\]

Here, $A$, $B$, and $C$ are constants we need to determine. To find these constants, we multiply both sides by the common denominator $(x + 1)(x^2 + 3x - 1)$ and simplify:

\[x - 5 = A(x^2 + 3x - 1) + (Bx + C)(x + 1)\]

Next, we strategically substitute values for $x$ to simplify the equations. Setting $x = -1$ eliminates the $A$ term:

\[-1 - 5 = A(1 - 3 - 1) \implies -6 = A(-3) \implies A = 2\]

Now, we substitute $A = 2$ back into the equation and expand:

\[x - 5 = 2(x^2 + 3x - 1) + (Bx + C)(x + 1)\]

Expanding gives:

\[x - 5 = 2x^2 + 6x - 2 + Bx^2 + (B + C)x + C\]

Combining like terms results in:

\[x - 5 = (2 + B)x^2 + (6 + B + C)x + (C - 2)\]

Setting coefficients equal, we have the following system of equations:

1. $2 + B = 0$ (for $x^2$ terms) 2. $6 + B + C = 1$ (for $x$ terms) 3. $C - 2 = -5$ (for constant terms)

From the first equation, we find $B = -2$. Substituting $B$ into the second equation:

\[6 - 2 + C = 1 \implies C = -3\]

Now we have $A = 2$, $B = -2$, and $C = -3$. Thus, we can rewrite the integral as:

\[\int_1^2 \left( \frac{2}{x + 1} - \frac{2x + 3}{x^2 + 3x - 1} \right) dx\]

We can now integrate each term separately. The first term integrates to:

\[2 \ln |x + 1|\]

For the second term, we use substitution. Let $u = x^2 + 3x - 1$, then $du = (2x + 3)dx$. The integral becomes:

\[-\int \frac{du}{u} = -\ln |u| = -\ln |x^2 + 3x - 1|\]

Combining these results, we have:

\[\int_1^2 \left( \frac{2}{x + 1} - \frac{2x + 3}{x^2 + 3x - 1} \right) dx = \left[ 2 \ln |x + 1| - \ln |x^2 + 3x - 1| \right]_1^2\]

Evaluating at the bounds:

For $x = 2$: \[2 \ln 3 - \ln 9\] For $x = 1$: \[2 \ln 2 - \ln 3\]

Thus, the definite integral evaluates to:

\[\left( 2 \ln 3 - \ln 9 \right) - \left( 2 \ln 2 - \ln 3 \right) = 2 \ln 3 - 2 \ln 2 + \ln 3 - \ln 9\]

Combining logarithmic terms gives:

\[\ln \left( \frac{3^3}{2^2 \cdot 9} \right) = \ln \left( \frac{27}{36} \right) = \ln \left( \frac{3}{4} \right)\]

Finally, the approximate value of the integral is:

\[\approx -0.29\]

This process illustrates the use of partial fraction decomposition and integration techniques to evaluate a definite integral of a rational function.

example

Partial Fraction Decomposition: Irreducible Quadratic Factors Example 6

Video duration:

22m

Partial Fraction Decomposition: Irreducible Quadratic Factors Example 6 Video Summary

To evaluate the integral of the rational function $\frac{5x^2 + 6}{(x^2 + 1)^2}$, we will utilize the method of partial fractions. This method is particularly useful when the degree of the numerator is less than that of the denominator, which is the case here. Since $x^2 + 1$ is an irreducible quadratic factor and is squared, we can express the integrand as a sum of simpler fractions.

We start by decomposing the function into partial fractions. The general form for our decomposition will be:

\[\frac{5x^2 + 6}{(x^2 + 1)^2} = \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2}\]

Next, we multiply through by the common denominator $(x^2 + 1)^2$ to eliminate the fractions:

\[5x^2 + 6 = (Ax + B)(x^2 + 1) + (Cx + D)\]

Expanding the right-hand side gives us:

\[5x^2 + 6 = Ax^3 + Bx^2 + Ax + B + Cx + D\]

Grouping like terms, we have:

\[5x^2 + 6 = Ax^3 + (B)x^2 + (A + C)x + (B + D)\]

To find the coefficients $A$, $B$, $C$, and $D$, we equate the coefficients from both sides of the equation:

For $x^3$: $A = 0$
For $x^2$: $B = 5$
For $x$: $A + C = 0$ (since $A = 0$, this implies $C = 0$)
For the constant term: $B + D = 6$ (substituting $B = 5$ gives $D = 1$)

Thus, we find $A = 0$, $B = 5$, $C = 0$, and $D = 1$. The partial fraction decomposition simplifies to:

\[\frac{5}{x^2 + 1} + \frac{1}{(x^2 + 1)^2}\]

Now we can integrate each term separately:

\[\int \left( \frac{5}{x^2 + 1} + \frac{1}{(x^2 + 1)^2} \right) dx = \int \frac{5}{x^2 + 1} dx + \int \frac{1}{(x^2 + 1)^2} dx\]

The first integral, $\int \frac{5}{x^2 + 1} dx$, can be solved using the known result:

\[\int \frac{1}{x^2 + 1} dx = \tan^{-1}(x)\]

Thus, we have:

\[\int \frac{5}{x^2 + 1} dx = 5 \tan^{-1}(x)\]

For the second integral, $\int \frac{1}{(x^2 + 1)^2} dx$, we can use a trigonometric substitution. Let $x = \tan(\theta)$, then $dx = \sec^2(\theta) d\theta$ and $x^2 + 1 = \sec^2(\theta)$. The integral becomes:

\[\int \frac{1}{\sec^4(\theta)} \sec^2(\theta) d\theta = \int \cos^2(\theta) d\theta\]

Using the identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$, we can rewrite the integral as:

\[\int \cos^2(\theta) d\theta = \frac{1}{2} \int (1 + \cos(2\theta)) d\theta = \frac{1}{2} \left( \theta + \frac{1}{2} \sin(2\theta) \right) + C\]

Substituting back for $\theta$ using $\theta = \tan^{-1}(x)$ and recognizing that $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$, we can express the final result in terms of $x$. The complete integral thus becomes:

\[5 \tan^{-1}(x) + \frac{1}{2} \tan^{-1}(x) + \frac{x}{2(x^2 + 1)} + C\]

Combining the results, we arrive at the final answer for the integral:

\[\frac{11}{2} \tan^{-1}(x) + \frac{x}{2(x^2 + 1)} + C\]

Problem

Use the method of partial fractions to evaluate the integral.
$\int_{}^{} \frac{5 x^{3} - x^{2} + 7 x - 2}{(x^{2} + 1) (x^{2} + 2)} d x$

$\ln ∣ x^{2} + 1 ∣ - \ln ∣ x^{2} + 1 ∣ + \frac{3}{2} \ln ∣ x^{2} + 2 ∣ + C$

$\ln ∣ x^{2} + 1 ∣ - t a n^{- 1} (x) + \frac{3}{2} \ln ∣ x^{2} + 2 ∣ + C$

$5 \ln ∣ x^{2} + 1 ∣ - \ln ∣ x^{2} + 1 ∣ + 3 \ln ∣ x^{2} + 2 ∣ + C$

$5 \ln ∣ x^{2} + 1 ∣ - t a n^{- 1} (x) + 3 \ln ∣ x^{2} + 2 ∣ + C$

Problem

Evaluate the integral.
$\int_{}^{} \frac{t^{4} + t^{2} - 2}{t^{3} + t} d t$

$\frac{t^{2}}{2} - 2 \ln ∣ t ∣ + \ln ∣ t^{2} + 1 ∣ + C$

$\frac{t^{2}}{2} - 2 \ln ∣ t ∣ + t a n^{- 1} (t) + C$

$\frac{t^{2}}{2} - 2 \ln ∣ t ∣ - t a n^{- 1} (t) + C$

$\frac{t^{2}}{2} - 2 \ln ∣ t ∣ - \ln ∣ t^{2} + 1 ∣ + C$

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Here’s what students ask on this topic:

What is partial fraction decomposition and why is it useful?

Partial fraction decomposition is a method used to break down a complex rational function into a sum of simpler fractions. This technique is particularly useful in calculus for integrating rational functions, as simpler fractions are easier to integrate. For example, a rational function with distinct linear factors can be expressed as constants over each factor. If the denominator includes repeated linear factors or irreducible quadratic factors, the decomposition adjusts accordingly, using terms for each power or linear expressions in the numerator. By simplifying the function, partial fraction decomposition allows for straightforward application of integration techniques, such as the natural logarithm rule or substitution methods.

How do you decompose a rational function with repeated linear factors?

To decompose a rational function with repeated linear factors, you must include terms for each power of the repeated factor up to its highest exponent. For example, if the denominator is ${(x + 1)}^{2}$ , the decomposition would include $\frac{A}{(x + 1)}$ and $\frac{B}{{(x + 1)}^{2}}$ . Each term corresponds to a power of the repeated factor. After setting up the fractions, multiply through by the common denominator and solve for constants using strategic substitutions or systems of equations. This ensures the rational function is fully decomposed for further analysis or integration.

What is the difference between distinct linear factors and irreducible quadratic factors in partial fraction decomposition?

Distinct linear factors are terms in the denominator that can be expressed as $ax + b$ , where the degree is one. For these, the numerator in the decomposition is a constant. Irreducible quadratic factors, on the other hand, are terms like ${ax}^{2} + bx + c$ that cannot be factored further. In this case, the numerator becomes a linear expression, such as $Ax + B$ . The decomposition process adjusts based on the type of factor, ensuring the rational function is expressed in a form suitable for integration or further simplification.

How do you solve for constants in partial fraction decomposition using strategic substitutions?

To solve for constants using strategic substitutions, choose values for $x$ that simplify the equation by eliminating terms. For example, if the denominator includes $(x - 2)$ , set $x = 2$ to make $(x - 2)$ zero, eliminating its associated term. Substitute $x$ into the equation and solve for the remaining constants. Repeat this process with other values of $x$ to find all constants. This method avoids expanding and grouping terms, making it faster and more efficient.

How do you integrate a rational function using partial fraction decomposition?

To integrate a rational function using partial fraction decomposition, first decompose the function into simpler fractions. For example, $\frac{f}{x}$ might be decomposed into $\frac{A}{(x + B)}$ and $\frac{C}{(x + D)}$ . Then, integrate each fraction separately. For terms like $\frac{1}{x}$ , use the natural logarithm rule: $\ln (x)$ . For irreducible quadratic factors, consider substitution or completing the square. Add the results and include the constant of integration $C$ . This method simplifies complex integrals into manageable steps.

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Partial Fractions: Videos & Practice Problems

Partial Fraction Decomposition: Distinct Linear Factors

Partial Fraction Decomposition: Distinct Linear Factors Video Summary

Partial Fraction Decomposition: Distinct Linear Factors Example 1

Partial Fraction Decomposition: Distinct Linear Factors Example 1 Video Summary

Express the rational function as a sum or difference of two simpler fractions. Use a system of equations.1(2x+1)(2x−1)\frac{1}{\left(2x+1\right)\left(2x-1\right)}

Express the rational function as a sum or difference of two simpler fractions. Use a system of equations.4x2−21x−40x3−x2−20x\frac{4x^2-21x-40}{x^3-x^2-20x}

Partial Fraction Decomposition: Distinct Linear Factors with Shortcut

Partial Fraction Decomposition: Distinct Linear Factors with Shortcut Video Summary

Give the partial fraction decomposition for the following expression using strategic substitutions for xx.4x2−8x−8x(x−2)(x+2)\frac{4x^2-8x-8}{x\left(x-2\right)\left(x+2\right)}

Express the rational function as a sum or difference or simpler rational expressions.4x3−3x2−x+3\frac{4}{x^3-3x^2-x+3}

Express the rational function as a sum or difference or simpler rational expressions.13x+22x2−x−1\frac{13x+2}{2x^2-x-1}

Integration Using Partial Fractions

Integration Using Partial Fractions Video Summary

Integration Using Partial Fractions Example 2

Integration Using Partial Fractions Example 2 Video Summary

Evaluate the integral. ∫−6x2+3x+5x3−xdx\int_{}^{}\frac{-6x^2+3x+5}{x^3-x}dx

Evaluate the integral. ∫033x+10x2+9x+20dx\int_0^3\frac{3x+10}{x^2+9x+20}dx

Evaluate the integral. ∫4x2+2x+3x2+xdx\int_{}^{}\frac{4x^2+2x+3}{x^2+x}dx

Find the area under the curve of y=10(x+2)(x+3)y=\frac{10}{(x+2)(x+3)} between x=0x=0 and x=4x=4.

Integration Using Partial Fractions Example 3

Integration Using Partial Fractions Example 3 Video Summary

Partial Fraction Decomposition: Repeated Linear Factors

Partial Fraction Decomposition: Repeated Linear Factors Video Summary

Partial Fraction Decomposition: Repeated Linear Factors Example 4

Partial Fraction Decomposition: Repeated Linear Factors Example 4 Video Summary

Express the integrand as a sum of partial fractions and evaluate the integral.∫2x3+x2−x−1dx\int_{}^{}\frac{2}{x^3+x^2-x-1}dx

Partial Fraction Decomposition: Repeated Linear Factors Example 5

Partial Fraction Decomposition: Repeated Linear Factors Example 5 Video Summary

Partial Fraction Decomposition: Irreducible Quadratic Factors

Partial Fraction Decomposition: Irreducible Quadratic Factors Video Summary

Partial Fraction Decomposition: Irreducible Quadratic Factors Example 6

Partial Fraction Decomposition: Irreducible Quadratic Factors Example 6 Video Summary

Partial Fraction Decomposition: Irreducible Quadratic Factors Example 6

Partial Fraction Decomposition: Irreducible Quadratic Factors Example 6 Video Summary

Use the method of partial fractions to evaluate the integral.∫5x3−x2+7x−2(x2+1)(x2+2)dx\int_{}^{}\frac{5x^3-x^2+7x-2}{(x^2+1)(x^2+2)}dx

Evaluate the integral.∫t4+t2−2t3+tdt\int_{}^{}\frac{t^4+t^2-2}{t^3+t}dt

Do you want more practice?

Here’s what students ask on this topic:

What is partial fraction decomposition and why is it useful?

How do you decompose a rational function with repeated linear factors?

What is the difference between distinct linear factors and irreducible quadratic factors in partial fraction decomposition?

How do you solve for constants in partial fraction decomposition using strategic substitutions?

How do you integrate a rational function using partial fraction decomposition?

Your Calculus tutors

Express the rational function as a sum or difference of two simpler fractions. Use a system of equations.
$\frac{1}{(2 x + 1) (2 x - 1)}$

Express the rational function as a sum or difference of two simpler fractions. Use a system of equations.
$\frac{4 x^{2} - 21 x - 40}{x^{3} - x^{2} - 20 x}$

Give the partial fraction decomposition for the following expression using strategic substitutions for $x$ .
$\frac{4 x^{2} - 8 x - 8}{x (x - 2) (x + 2)}$

Express the rational function as a sum or difference or simpler rational expressions.
$\frac{4}{x^{3} - 3 x^{2} - x + 3}$

Express the rational function as a sum or difference or simpler rational expressions.
$\frac{13 x + 2}{2 x^{2} - x - 1}$

Evaluate the integral.
$\int_{}^{} \frac{- 6 x^{2} + 3 x + 5}{x^{3} - x} d x$

Evaluate the integral.
$\int_{0}^{3} \frac{3 x + 10}{x^{2} + 9 x + 20} d x$

Evaluate the integral.
$\int_{}^{} \frac{4 x^{2} + 2 x + 3}{x^{2} + x} d x$

Find the area under the curve of $y = \frac{10}{(x + 2) (x + 3)}$ between $x equals 0$ and $x equals 4$ .

Express the integrand as a sum of partial fractions and evaluate the integral.
$\int_{}^{} \frac{2}{x^{3} + x^{2} - x - 1} d x$

Use the method of partial fractions to evaluate the integral.
$\int_{}^{} \frac{5 x^{3} - x^{2} + 7 x - 2}{(x^{2} + 1) (x^{2} + 2)} d x$

Evaluate the integral.
$\int_{}^{} \frac{t^{4} + t^{2} - 2}{t^{3} + t} d t$