Find the integral. ∫ s e 3 s d s \int_{}^{}se^{3s}ds ∫ s e 3 s d s

s 3 e 3 s − 1 9 e 3 s + C \frac{s}{3}e^{3s}-\frac19e^{3s}+C 3 s e 3 s − 9 1 e 3 s + C

s 9 e 3 s − 1 3 e 3 s + C \frac{s}{9}e^{3s}-\frac13e^{3s}+C

3 s e s − 1 9 e 3 s + C 3se^{s}-\frac19e^{3s}+C 3 s e s − 9 1 e 3 s + C

3 s e 3 s − 9 e 3 s + C 3se^{3s}-9e^{3s}+C 3 s e 3 s − 9 e 3 s + C

Find the integral. ∫ x 2 ln ⁡ x d x \int_{}^{}x^2\ln xdx ∫ x 2 ln x d x

x 3 3 ln ⁡ x − x 3 9 + C \frac{x^3}{3}\ln x-\frac{x^3}{9}+C 3 x 3 ln x − 9 x 3 + C

x 3 3 ln ⁡ x + x 3 9 + C \frac{x^3}{3}\ln x+\frac{x^3}{9}+C 3 x 3 ln x + 9 x 3 + C

x 3 3 ln ⁡ x + 1 x + C \frac{x^3}{3}\ln x+\frac{1}{x}+C 3 x 3 ln x + x 1 + C

x 3 3 ln ⁡ x − 1 x + C \frac{x^3}{3}\ln x-\frac{1}{x}+C 3 x 3 ln x − x 1 + C

Find the indefinite integral. ∫ r 2 e − r d r \int_{}^{}r^2e^{-r}dr ∫ r 2 e − r d r

− r 2 e − r − 2 r e − r − 2 e − r + C -r^2e^{-r}-2re^{-r}-2e^{-r}+C − r 2 e − r − 2 r e − r − 2 e − r + C

r 2 e − r − 2 r e − r + 2 e − r + C r^2e^{-r}-2re^{-r}+2e^{-r}+C r 2 e − r − 2 r e − r + 2 e − r + C

− 1 2 r 2 e − r − r e − r − e − r + C -\frac12r^2e^{-r}-re^{-r}-e^{-r}+C − 2 1 r 2 e − r − r e − r − e − r + C

1 2 r 2 e − r − r e − r + e − r + C \frac12r^2e^{-r}-re^{-r}+e^{-r}+C 2 1 r 2 e − r − r e − r + e − r + C

Evaluate the indefinite integral. ∫ ( s 2 + 4 ) e 3 s d s \int_{}^{}\left(s^2+4\right)e^{3s}ds ∫ ( s 2 + 4 ) e 3 s d s

e 3 s 3 ( s 2 + 4 ) − 2 9 s e 3 s + C \frac{e^{3s}}{3}\left(s^2+4\right)-\frac29se^{3s}+C 3 e 3 s ( s 2 + 4 ) − 9 2 s e 3 s + C

( s 3 + 4 3 s ) e 3 s + C \left(s^3+\frac43s\right)e^{3s}+C ( s 3 + 3 4 s ) e 3 s + C

( s 3 3 + 4 3 s ) e 3 s + C \left(\frac{s^3}{3}+\frac43s\right)e^{3s}+C ( 3 s 3 + 3 4 s ) e 3 s + C

Evaluate the indefinite integral. ∫ ln ⁡ x x 4 d x \int^{}\frac{\ln x}{x^4}dx ∫ x 4 l n x d x

− ln ⁡ x 3 x 3 − 1 9 x 3 + C -\frac{\ln x}{3x^3}-\frac{1}{9x^3}+C − 3 x 3 l n x − 9 x 3 1 + C

− ln ⁡ x 3 x 3 + 1 x + C -\frac{\ln x}{3x^3}+\frac{1}{x}+C − 3 x 3 l n x + x 1 + C

ln ⁡ x 3 x 3 + 1 x + C \frac{\ln x}{3x^3}+\frac{1}{x}+C 3 x 3 l n x + x 1 + C

ln ⁡ x 3 x 3 − 1 9 x 3 + C \frac{\ln x}{3x^3}-\frac{1}{9x^3}+C 3 x 3 l n x − 9 x 3 1 + C

11. Techniques of Integration

Integration by Parts

11. Techniques of Integration

Integration by Parts: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Integration by parts is a powerful technique for solving integrals involving the product of two functions, especially when variable substitution is ineffective. The formula $\int u^{d} v = u \cdot v - \int v d u$ allows for systematic integration, often requiring repetition for complex cases. For definite integrals, apply bounds to both parts of the equation after integration. Mastery of this method enhances problem-solving skills in calculus.

concept

Introduction to Integration by Parts

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Introduction to Integration by Parts Video Summary

When faced with the integral of a product of two functions, such as $ \int 6x e^{x^2} \, dx $, a common approach is to use variable substitution. However, if the problem changes to $ \int 6x e^x \, dx $, this method becomes ineffective. In such cases, integration by parts (IBP) is a valuable technique to apply.

Integration by parts is based on the product rule of differentiation, which states that the derivative of the product of two functions $ f(x) $ and $ g(x) $ can be expressed as:

\[\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)\]

To derive the integration by parts formula, we can integrate both sides with respect to $ x $, leading to:

\[\int f'(x)g(x) \, dx = f(x)g(x) - \int f(x)g'(x) \, dx\]

Rearranging this gives us the integration by parts formula:

\[\int u \, dv = uv - \int v \, du\]

In this formula, $ u $ is a function that simplifies when differentiated, and $ dv $ is a function that can be easily integrated to find $ v $.

To apply this to our example $ \int 6x e^x \, dx $, we can choose:

Let $ u = 6x $ (which simplifies to $ du = 6 \, dx $)
Let $ dv = e^x \, dx $ (which integrates to $ v = e^x $)

Substituting these into the integration by parts formula gives:

\[\int 6x e^x \, dx = 6x e^x - \int e^x \cdot 6 \, dx\]

This simplifies to:

\[6x e^x - 6 \int e^x \, dx\]

Since $ \int e^x \, dx = e^x + C $, we can substitute this back into our equation:

\[6x e^x - 6e^x + C\]

Thus, the final result for the integral $ \int 6x e^x \, dx $ is:

\[6x e^x - 6e^x + C\]

Understanding how to identify $ u $ and $ dv $ is crucial for successfully applying integration by parts. Practice with various examples will help solidify this technique, making it easier to tackle more complex integrals in the future.

example

Introduction to Integration by Parts Example 1

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Introduction to Integration by Parts Example 1 Video Summary

To solve the integral of the natural logarithm of $ x $ with respect to $ x $, we can utilize the method of integration by parts. This technique is particularly useful when we do not have a direct integral formula for the function we are working with. The integration by parts formula is given by:

\[\int u \, dv = uv - \int v \, du\]

In this case, we will choose $ u = \ln(x) $ because its derivative, $ du = \frac{1}{x} \, dx $, simplifies the expression. For $ dv $, we take $ dv = dx $, which is straightforward to integrate, yielding $ v = x $.

Now, substituting these values into the integration by parts formula, we have:

\[\int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx\]

Notice that the $ x $ in the numerator and denominator cancels out, simplifying our integral to:

\[\int \ln(x) \, dx = x \ln(x) - \int 1 \, dx\]

The integral of $ 1 $ with respect to $ x $ is simply $ x $. Therefore, we can express the solution as:

\[\int \ln(x) \, dx = x \ln(x) - x + C\]

where $ C $ is the constant of integration. This result shows how integration by parts can effectively handle the integral of the natural logarithm.

Additionally, a helpful mnemonic for choosing $ u $ and $ dv $ is the acronym "LIATE," which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions. According to this rule, logarithmic functions should be chosen as $ u $ when present, as they simplify upon differentiation.

Problem

Find the integral.
$\int_{}^{}se^{3s}ds$

$\frac{s}{9} e^{3 s} - \frac{1}{3} e^{3 s} + C$

$3se^{s}-\frac19e^{3s}+C$

$\frac{s}{3}e^{3s}-\frac19e^{3s}+C$

$3se^{3s}-9e^{3s}+C$

Problem

Find the integral.
$\int_{}^{}x^2\ln xdx$

$\frac{x^3}{3}\ln x+\frac{x^3}{9}+C$

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

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

$\frac{x^3}{3}\ln x-\frac{x^3}{9}+C$

concept

Repeated Integration by Parts

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Repeated Integration by Parts Video Summary

Integration by parts is a powerful technique used to evaluate integrals, particularly when dealing with the product of two functions. This method is especially useful when other strategies, such as variable substitution, are not applicable. The fundamental formula for integration by parts is given by:

$$\int u \, dv = uv - \int v \, du$$

In this context, selecting the appropriate functions for $u$ and $dv$ is crucial. The function $u$ should be chosen such that its derivative $du$ simplifies the integral, while $dv$ should be a function that is easy to integrate. For example, when evaluating the integral of $3x^2 e^x$, we can set $u = 3x^2$ and $dv = e^x \, dx$. This choice leads to:

$$du = 6x \, dx \quad \text{and} \quad v = e^x$$

Applying the integration by parts formula, we have:

$$\int 3x^2 e^x \, dx = 3x^2 e^x - \int e^x (6x) \, dx$$

At this point, we encounter a new integral, $\int 6x e^x \, dx$, which requires us to apply integration by parts again. Here, we can set $u = 6x$ and $dv = e^x \, dx$, leading to:

$$du = 6 \, dx \quad \text{and} \quad v = e^x$$

Thus, we can express the integral as:

$$\int 6x e^x \, dx = 6x e^x - \int e^x (6) \, dx$$

Now, integrating $e^x$ is straightforward, yielding:

$$\int e^x \, dx = e^x$$

Combining all parts, we arrive at the final expression:

$$\int 3x^2 e^x \, dx = 3x^2 e^x - 6x e^x + 6e^x + C$$

In summary, integration by parts may require multiple applications to solve more complex integrals. The key is to recognize when to repeat the process and to maintain clarity in your selections for $u$ and $dv$. Mastering this technique will enhance your ability to tackle a wide range of integration problems effectively.

example

Repeated Integration by Parts Example 2

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Repeated Integration by Parts Example 2 Video Summary

To solve the indefinite integral of the product of two functions, we can utilize the integration by parts formula, which states:

Integration by Parts Formula: \[\int u \, dv = uv - \int v \, du\]

In this case, we have a polynomial and an exponential function. We start by selecting $ u $ as the polynomial $ t^2 + 2t - 3 $, since its derivative will simplify the expression. The derivative $ du $ is calculated as follows:

\[du = (2t + 2) \, dt\]

Next, we identify $ dv $ as $ e^{4t} \, dt $. Integrating this gives us:

\[v = \frac{1}{4} e^{4t}\]

Now we can substitute $ u $, $ du $, $ v $, and $ dv $ into the integration by parts formula:

\[\int (t^2 + 2t - 3)e^{4t} \, dt = (t^2 + 2t - 3) \cdot \frac{1}{4} e^{4t} - \int \frac{1}{4} e^{4t} (2t + 2) \, dt\]

We can factor out $ \frac{1}{4} $ from the integral:

\[= \frac{1}{4} (t^2 + 2t - 3)e^{4t} - \frac{1}{4} \int e^{4t} (2t + 2) \, dt\]

To evaluate the remaining integral, we apply integration by parts again. This time, we set $ u = 2t + 2 $ and $ dv = e^{4t} \, dt $. The derivative $ du $ is:

\[du = 2 \, dt\]

And integrating $ dv $ gives us the same $ v $ as before:

\[v = \frac{1}{4} e^{4t}\]

Substituting these into the integration by parts formula again yields:

\[\int (2t + 2)e^{4t} \, dt = (2t + 2) \cdot \frac{1}{4} e^{4t} - \int \frac{1}{4} e^{4t} \cdot 2 \, dt\]

Factoring out $ \frac{1}{4} $ gives:

\[= \frac{1}{4} (2t + 2)e^{4t} - \frac{1}{2} \int e^{4t} \, dt\]

The integral of $ e^{4t} $ is straightforward:

\[\int e^{4t} \, dt = \frac{1}{4} e^{4t}\]

Substituting this back, we have:

\[= \frac{1}{4} (2t + 2)e^{4t} - \frac{1}{2} \cdot \frac{1}{4} e^{4t} = \frac{1}{4} (2t + 2)e^{4t} - \frac{1}{8} e^{4t}\]

Now, we can combine everything back into our original expression:

\[\int (t^2 + 2t - 3)e^{4t} \, dt = \frac{1}{4} (t^2 + 2t - 3)e^{4t} - \frac{1}{4} \left( \frac{1}{4} (2t + 2)e^{4t} - \frac{1}{8} e^{4t} \right)\]

After simplifying, we arrive at the final expression for the indefinite integral:

\[= \frac{1}{4} (t^2 + 2t - 3)e^{4t} - \frac{1}{16} (2t + 2)e^{4t} + \frac{1}{32} e^{4t} + C\]

In conclusion, the process of repeated integration by parts allows us to break down complex integrals into simpler components, ultimately leading to the solution. The key is to choose $ u $ wisely so that its derivative simplifies the integral.

example

Repeated Integration by Parts Example 3

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Repeated Integration by Parts Example 3 Video Summary

To find the indefinite integral of the natural logarithm of $x^2$, we will utilize integration by parts, a technique particularly useful when dealing with logarithmic functions. The formula for integration by parts is given by:

\[\int u \, dv = uv - \int v \, du\]

In this case, we set $u = \ln(x^2)$ and differentiate it to find $du$. Using the chain rule, we have:

\[du = \frac{d}{dx}(\ln(x^2)) = \frac{2}{x} \, dx\]

Next, we choose $dv = dx$, which integrates to $v = x$. Now we can apply the integration by parts formula:

\[\int \ln(x^2) \, dx = x \ln(x^2) - \int x \cdot \frac{2}{x} \, dx\]

Notice that the $x$ terms cancel, simplifying our integral to:

\[\int \ln(x^2) \, dx = x \ln(x^2) - 2 \int dx\]

Integrating $dx$ gives us $x$, leading to:

\[\int \ln(x^2) \, dx = x \ln(x^2) - 2x + C\]

Next, we simplify $x \ln(x^2)$ using the property of logarithms, $\ln(x^2) = 2 \ln(x)$, resulting in:

\[\int \ln(x^2) \, dx = 2x \ln(x) - 2x + C\]

Thus, the final result for the indefinite integral of $\ln(x^2)$ is:

\[\int \ln(x^2) \, dx = 2x \ln(x) - 2x + C\]

This process illustrates the effectiveness of integration by parts, especially when the function simplifies with each differentiation. Repeating the integration by parts method can be tedious, but it is essential for solving integrals involving logarithmic functions.

Problem

Find the indefinite integral.
$\int_{}^{}r^2e^{-r}dr$

$r^2e^{-r}-2re^{-r}+2e^{-r}+C$

$-r^2e^{-r}-2re^{-r}-2e^{-r}+C$

$-\frac12r^2e^{-r}-re^{-r}-e^{-r}+C$

$\frac12r^2e^{-r}-re^{-r}+e^{-r}+C$

concept

Tabular Integration by Parts

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Tabular Integration by Parts Video Summary

Integration by parts is a powerful technique used to solve integrals involving the product of two functions. The fundamental formula for integration by parts is given by:

$$\int u \, dv = uv - \int v \, du$$

In this context, choosing the right functions for $ u $ and $ dv $ is crucial. A helpful mnemonic for selecting $ u $ is LIATE, which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions. Typically, you want to choose $ u $ from the algebraic category when dealing with polynomials, as they simplify upon differentiation.

For more complex integrals, the tabular method can streamline the process. This method organizes the derivatives and integrals into a table, making it easier to visualize the steps involved. To illustrate, consider the integral of $ 3x^2 e^{2x} $. First, identify $ u $ and $ dv $: set $ u = 3x^2 $ and $ dv = e^{2x} \, dx $. Then, compute the derivatives of $ u $ and the integrals of $ dv $.

For the derivatives:

First derivative: $ \frac{d}{dx}(3x^2) = 6x $
Second derivative: $ \frac{d}{dx}(6x) = 6 $
Third derivative: $ \frac{d}{dx}(6) = 0 $

For the integrals:

First integral: $ \int e^{2x} \, dx = \frac{1}{2} e^{2x} $
Second integral: $ \int \frac{1}{2} e^{2x} \, dx = \frac{1}{4} e^{2x} $
Third integral: $ \int \frac{1}{4} e^{2x} \, dx = \frac{1}{8} e^{2x} $

Next, arrange these results in a table with alternating signs. The first column contains the derivatives of $ u $, and the second column contains the integrals of $ dv $. The pattern of signs starts with a positive sign for the first term, followed by negative, positive, and so on.

Using the table, the integral can be computed as follows:

$$\int 3x^2 e^{2x} \, dx = 3x^2 \cdot \frac{1}{2} e^{2x} - 6x \cdot \frac{1}{4} e^{2x} + 6 \cdot \frac{1}{8} e^{2x} + C$$

After simplifying, the final result is:

$$\int 3x^2 e^{2x} \, dx = \frac{3}{2} x^2 e^{2x} - \frac{3}{2} x e^{2x} + \frac{3}{4} e^{2x} + C$$

In summary, the tabular method for integration by parts not only organizes the process but also reduces the complexity of repeated applications of the integration by parts formula. This method is particularly useful when dealing with higher-order polynomials and exponential functions, allowing for a more efficient calculation of integrals.

example

Tabular Integration by Parts Example 4

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Tabular Integration by Parts Example 4 Video Summary

To evaluate the indefinite integral using the tabular method, we start by recognizing that integration by parts is necessary when dealing with the product of functions. The tabular method simplifies this process by organizing derivatives and integrals into a table format. In this case, we need to set up two columns: one for derivatives (denoted as the d column) and another for integrals (denoted as the I column).

Choosing the appropriate u and dv is crucial. The goal is to select u such that its derivative simplifies the expression. In this scenario, we set u as the natural logarithm, ln(t), because its derivative, 1/t, is simpler than the alternative choice of t³. The dv is then set as t³ dt, which can be easily integrated using the power rule.

The derivatives and integrals are organized as follows:

d column: ln(t) (1st derivative: 1/t)
I column: t³ (integral: t⁴/4)

Next, we apply the integration by parts formula, which involves multiplying the entries diagonally and integrating the remaining terms. The first term from the table gives us:

ln(t) * (t⁴/4)

For the second term, we integrate the product of the remaining entries, which leads to:

- ∫(1/t * t⁴/4) dt

This simplifies to:

- ∫(t³/4) dt

Integrating t³ yields t⁴/4, allowing us to combine the results. The final expression becomes:

ln(t) * (t⁴/4) - (t⁴/16) + C

In conclusion, the indefinite integral evaluated using the tabular method results in:

ln(t) * (t⁴/4) - (t⁴/16) + C

This method effectively streamlines the integration process, particularly for products of functions, by organizing the necessary components into a clear and manageable format.

Problem

Evaluate the indefinite integral.
$\int_{}^{}\left(s^2+4\right)e^{3s}ds$

$\frac{e^{3s}}{3}\left(s^2+4\right)-\frac29se^{3s}+\frac{2}{27}e^{3s}+C$

$\frac{e^{3s}}{3}\left(s^2+4\right)-\frac29se^{3s}+C$

$\left(s^3+\frac43s\right)e^{3s}+C$

$\left(\frac{s^3}{3}+\frac43s\right)e^{3s}+C$

Problem

Evaluate the indefinite integral.
$\int^{}\frac{\ln x}{x^4}dx$

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

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

$\frac{\ln x}{3x^3}-\frac{1}{9x^3}+C$

$-\frac{\ln x}{3x^3}-\frac{1}{9x^3}+C$

concept

Integration by Parts for Definite Integrals

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Integration by Parts for Definite Integrals Video Summary

Integration by parts is a powerful technique used to evaluate integrals, and it can be applied to both indefinite and definite integrals. The fundamental formula for integration by parts states that the integral of $ u \, dv $ can be expressed as:

$ \int u \, dv = uv - \int v \, du $

When dealing with definite integrals, the process remains largely the same, but with an important adjustment: after applying the integration by parts formula, you must evaluate both $ u \times v $ and the integral $ \int v \, du $ at the specified bounds. This means that you will apply the limits of integration to both parts of the equation, which is a common point of confusion.

To illustrate this, consider a definite integral where you choose $ u $ and $ dv $. For example, if you let $ u = 6x $ and $ dv = e^x \, dx $, then the derivatives and integrals are $ du = 6 \, dx $ and $ v = e^x $. Applying the integration by parts formula gives:

$ \int_0^2 6x \, e^x \, dx = [6x \cdot e^x]_0^2 - \int_0^2 e^x \cdot 6 \, dx $

Next, you evaluate the first term at the bounds $ 0 $ and $ 2 $. This results in:

$ [6(2) \cdot e^2 - 6(0) \cdot e^0] = 12e^2 - 0 = 12e^2 $

For the integral $ \int_0^2 6e^x \, dx $, you can factor out the constant $ 6 $ and integrate $ e^x $, yielding:

$ 6[e^x]_0^2 = 6(e^2 - e^0) = 6(e^2 - 1) $

Combining these results, you have:

$ 12e^2 - 6(e^2 - 1) = 12e^2 - 6e^2 + 6 = 6e^2 + 6 $

This can be factored to give the final result:

$ 6(e^2 + 1) $

If a decimal approximation is required, you can compute this using a calculator, which would yield approximately $ 50.33 $. This example demonstrates how to effectively apply integration by parts to definite integrals, ensuring that you remember to evaluate both components at the specified bounds.

Problem

Evaluate the definite integral.
$\int_1^2x\ln xdx$

$-0.11$

$0.64$

$0.89$

$1.89$

Problem

The rate of growth of a particular bacteria is given by $b^{'} (t) = 4 t^{2} e^{.5 t}$ where $t$ is time in days. What is the total growth of the population of bacteria during the first 5 days?

3,216

$1 comma 656$

$1 comma 203$

$2 comma 826$

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We have more practice problems on Integration by Parts

Here’s what students ask on this topic:

What is the formula for integration by parts and how is it derived?

The formula for integration by parts is derived from the product rule for differentiation. The formula is:

\int u d v = u v - \int v d u

To derive it, start with the product rule for differentiation:

(u v)' = u v' + v u'

Integrate both sides with respect to x:

\int (u v)' d x = \int u v' d x + \int v u' d x

Rearranging terms gives the integration by parts formula. This method is useful for integrating products of functions where substitution fails.

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How do you choose u and dv in integration by parts?

Choosing u and dv in integration by parts is crucial for simplifying the integral. A common guideline is the LIATE rule, which prioritizes functions in the following order:

L: Logarithmic functions (e.g., ln(x))
I: Inverse trigonometric functions (e.g., arctan(x))
A: Algebraic functions (e.g., x, x²)
T: Trigonometric functions (e.g., sin(x), cos(x))
E: Exponential functions (e.g., e^x)

Choose u as the function that simplifies when differentiated, and dv as the function that is easy to integrate. For example, in the integral of x·e^x, set u = x (derivative simplifies to 1) and dv = e^xdx (integral remains e^x).

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What is the tabular method for integration by parts?

The tabular method is a systematic way to apply integration by parts multiple times, especially for functions requiring repeated differentiation and integration. Create two columns: one for derivatives (u) and one for integrals (dv). Differentiate u until it reaches zero, and integrate dv the same number of times. Alternate signs (+, −, +, ...) as you multiply diagonally across the table. For example, to integrate 3x²·e^x, set u = 3x² and dv = e^xdx. Differentiate u to 0 and integrate dv repeatedly. Multiply diagonally and sum the terms to find the solution.

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How do you apply integration by parts to definite integrals?

To apply integration by parts to definite integrals, use the formula:

\int a_{b} u d v = a_{b} u v - \int a_{b} v d u

First, solve the integral as if it were indefinite. Then, apply the bounds to both u·v and the remaining integral. For example, to evaluate:

\int 0_{2} 6 x e^x d x

Set u = 6x and dv = e^xdx. After integration, apply bounds to both terms. Subtract the lower bound result from the upper bound result to find the definite integral.

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What are common mistakes to avoid when using integration by parts?

Common mistakes in integration by parts include:

Incorrect choice of u and dv: Ensure u simplifies when differentiated and dv is easy to integrate.
Forgetting the constant of integration: Always add +C for indefinite integrals.
Misapplying bounds: For definite integrals, apply bounds to both u·v and the remaining integral.
Skipping steps: Carefully compute derivatives and integrals to avoid errors.
Not repeating the process: Some integrals require multiple applications of the formula.