Find the indefinite integral. ∫ 3 − y 2 2 y d y \int_{}^{}\frac{3-y^2}{2y}dy ∫ 2 y 3 − y 2 d y

3 2 ln ⁡ ∣ y ∣ − 1 4 y 2 + C \frac32\ln\left|y\right|-\frac14y^2+C 2 3 ln ∣ y ∣ − 4 1 y 2 + C

3 2 + 1 4 y 2 + C \frac32+\frac{1}{4y^2}+C 2 3 + 4 y 2 1 + C

3 2 − 1 4 y 2 + C \frac32-\frac{1}{4y^2}+C 2 3 − 4 y 2 1 + C

3 2 ln ⁡ ∣ y ∣ − 1 4 y 2 + C \frac32\ln\left|y\right|-\frac{1}{4y^2}+C 2 3 ln ∣ y ∣ − 4 y 2 1 + C

What is the integral of 1/x and how is it derived?

The integral of 1/x is derived from reversing the process of differentiation. The derivative of the natural logarithm of x, ln(x), is 1/x. Therefore, when we integrate 1/x, we reverse this process and obtain the natural logarithm of the absolute value of x, plus a constant of integration. The integral is expressed as: ∫ 1 x d x = ln | x | + c Here, the absolute value ensures the function is defined for all x values, including negative ones. This integral is fundamental in calculus and is used in various applications, including solving differential equations and evaluating definite integrals.

When should you use the power rule versus the natural log rule in integration?

Choosing between the power rule and the natural log rule in integration depends on the form of the function. Use the power rule when integrating functions of the form x n , where n is not -1. The power rule states: ∫ x n d x = x n + 1 n + 1 + c Use the natural log rule when integrating 1/x, as its integral is ln|x| + c. Recognizing the correct rule is essential for solving integrals effectively, as applying the wrong rule can lead to incorrect results.

How do you integrate a function like 1/x 2 ?

To integrate a function like 1/x 2 , rewrite it using a negative exponent: x -2 . Then, apply the power rule for integration, which is suitable for powers other than -1. The integral is: ∫ x - 2 d x = - 1 x + c Here, you add 1 to the exponent (-2 + 1 = -1) and divide by the new exponent (-1), resulting in -1/x + c. This approach is different from integrating 1/x, which results in a natural logarithm.

10. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

10. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions: Videos & Practice Problems

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Topic summary

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Understanding the integration of $\frac{1}{x}$ is crucial. The integral of $\frac{1}{x}$ is the natural logarithm of the absolute value of x, plus a constant: $\ln | x | + c$ . This rule is derived from reversing the derivative process. Constants can be factored out, simplifying integration. For example, integrating $\frac{5}{x}$ results in $5 \ln | x | + c$ . Use the power rule for terms like $\frac{1}{x^{2}}$ , rewriting as $x^{- 2}$ .

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concept

Integrals Resulting in Natural Logs

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Integrals Resulting in Natural Logs Video Summary

In calculus, understanding the relationship between derivatives and integrals is crucial. The derivative of the natural logarithm function, specifically ln(x), is given by the formula:

\[\frac{d}{dx} \ln(x) = \frac{1}{x}\]

This relationship allows us to derive a corresponding integral rule. By reversing the differentiation process, we can find the integral of 1/x. Thus, the integral can be expressed as:

\[\int \frac{1}{x} \, dx = \ln |x| + C\]

Here, C represents the constant of integration, and the absolute value is included to ensure the function is defined for all values of x, including negative values.

Additionally, 1/x can be rewritten as x^-1, which emphasizes that negative exponents indicate reciprocal relationships. This understanding is essential when applying the constant multiple rule in integration.

For example, when integrating 5/x, we can factor out the constant:

\[\int \frac{5}{x} \, dx = 5 \int \frac{1}{x} \, dx = 5 \ln |x| + C\]

Next, consider a more complex integral, such as:

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

This can be separated into two distinct integrals using the sum rule:

\[\int \frac{1}{x^2} \, dx + \int \frac{3}{x} \, dx\]

For the first integral, 1/x² can be rewritten as x^-2. Applying the power rule for integration, we have:

\[\int x^{-2} \, dx = -\frac{1}{x}\]

For the second integral, we again apply the constant multiple rule:

\[\int \frac{3}{x} \, dx = 3 \int \frac{1}{x} \, dx = 3 \ln |x|\]

Combining these results, the final answer for the integral becomes:

\[-\frac{1}{x} + 3 \ln |x| + C\]

Through these examples, we see the application of integration rules, including the power rule and the constant multiple rule, reinforcing the foundational concepts of calculus. Continued practice with various integrals will further enhance understanding and proficiency in this area.

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example

Integrals Resulting in Natural Logs Example 1

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Integrals Resulting in Natural Logs Example 1 Video Summary

To solve the indefinite integral of the expression $ \frac{x^3 e^x - 4x^2}{x^3} \, dx $, we can simplify the problem by breaking it down into two separate fractions. This allows us to handle each term more easily.

First, we rewrite the integral as follows:

\[\int \left( \frac{x^3 e^x}{x^3} - \frac{4x^2}{x^3} \right) \, dx\]

Upon simplifying, we notice that the $ x^3 $ in the numerator cancels with the $ x^3 $ in the denominator for the first term, and in the second term, $ x^2 $ cancels with $ x^3 $, leaving us with $ x $ in the denominator:

\[\int \left( e^x - \frac{4}{x} \right) \, dx\]

This integral is now much simpler to evaluate. The antiderivative of $ e^x $ is simply $ e^x $. For the second term, we apply the rule for integrating $ \frac{1}{x} $, which gives us $ -4 \ln |x| $. Therefore, we can combine these results:

\[\int \left( e^x - \frac{4}{x} \right) \, dx = e^x - 4 \ln |x| + C\]

Here, $ C $ represents the constant of integration. Thus, the final answer for the indefinite integral is:

\[e^x - 4 \ln |x| + C\]

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Problem

Find the indefinite integral.
$\int_{}^{}\frac{3-y^2}{2y}dy$

$\frac32\ln\left|y\right|-\frac14y^2+C$

$\frac32+\frac{1}{4y^2}+C$

$\frac32-\frac{1}{4y^2}+C$

$\frac32\ln\left|y\right|-\frac{1}{4y^2}+C$

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The integral of 1/x is derived from reversing the process of differentiation. The derivative of the natural logarithm of x, ln(x), is 1/x. Therefore, when we integrate 1/x, we reverse this process and obtain the natural logarithm of the absolute value of x, plus a constant of integration. The integral is expressed as:

\int \frac{1}{x} d x = \ln | x | + c

Here, the absolute value ensures the function is defined for all x values, including negative ones. This integral is fundamental in calculus and is used in various applications, including solving differential equations and evaluating definite integrals.

To integrate a function like 5/x, you can use the constant multiple rule, which allows you to factor out constants from the integral. The integral of 5/x is:

\int \frac{5}{x} d x = 5 \int \frac{1}{x} d x

Using the rule for integrating 1/x, this becomes:

5 \ln | x | + c

Here, ln|x| is the natural logarithm of the absolute value of x, and c is the constant of integration. This process simplifies the integration by handling the constant separately, making it easier to solve.

Choosing between the power rule and the natural log rule in integration depends on the form of the function. Use the power rule when integrating functions of the form xⁿ, where n is not -1. The power rule states:

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c

Use the natural log rule when integrating 1/x, as its integral is ln|x| + c. Recognizing the correct rule is essential for solving integrals effectively, as applying the wrong rule can lead to incorrect results.

To integrate a function like 1/x², rewrite it using a negative exponent: x^-2. Then, apply the power rule for integration, which is suitable for powers other than -1. The integral is:

\int x^{- 2} d x = \frac{-}{1} + c

Here, you add 1 to the exponent (-2 + 1 = -1) and divide by the new exponent (-1), resulting in -1/x + c. This approach is different from integrating 1/x, which results in a natural logarithm.

The constant of integration, denoted as c, is crucial in indefinite integrals because it represents the family of antiderivatives. When you integrate a function, you find one of many possible antiderivatives, as differentiation removes constant terms. The constant of integration accounts for this lost information, ensuring that all possible antiderivatives are represented. For example, the integral of 1/x is ln|x| + c, where c can be any constant. This constant is important in applications such as solving differential equations, where initial conditions determine the specific value of c. Without the constant of integration, the solution would be incomplete, potentially leading to incorrect results.