# Integration by Parts – Examples with Answers

Integration by parts allows us to “reduce” an integral to a simpler form, expressing it as the difference between two simpler integrals. This technique is especially useful when we want to evaluate integrals that cannot be easily found using other methods, such as substitution or trigonometric identities.

In this article, we will look at some solved integration by parts exercises. Then, we will see some practice problems to apply what we have learned.

##### CALCULUS

Relevant for

Learning about integration by parts with examples.

See examples

##### CALCULUS

Relevant for

Learning about integration by parts with examples.

See examples

## How to integrate functions by parts

Integration by parts is used to integrate the product of two functions. To integrate functions using this method, we follow the steps below:

#### 1. Choose two functions, u and dv/dx

The product of the two functions, $latex u\frac{dv}{dx}$ is the integrand.

#### 4. Use the formula for integration by parts:

$$\int u \frac{dv}{dx} dx=uv – \int v \frac{du}{dx}dx$$

or

$$\int u v^{\prime} dx=uv – \int v u^{\prime}dx$$

When using integration by parts, the choice of $latex u$ and $latex dv$ is not always obvious. However, there are some rules of thumb that can help guide your choice.

First, $latex u$ should be chosen such that it is an easy function to integrate, while $latex dv$ should be chosen such that it is an easy function to differentiate. This will make the integration and differentiation steps in the integration by parts formula simpler and easier to evaluate.

Second, it can sometimes be useful to choose $latex u$ and $latex dv$ so that the product $latex uv$ is as simple as possible. This can make the final step of the integration by parts formula easier to evaluate since you will be left with an integral that has a simple integrand.

## Integration by parts – Examples with answers

### EXAMPLE 1

Find the following integral:

$$\int x\cos x dx$$

To solve this integral, in which the product of two functions appears, the rule of integration by parts is used, according to which:

$$\int f(x)g'(x)dx=f(x)g(x)-\int g(x)f'(x)dx$$

If we specify the following:

• $latex f(x)=u$
• $latex g'(x)dx=dv\Rightarrow v=g(x)$

Then, the expression at the beginning is rewritten as follows:

$$\int udv=uv -\int vdu$$

Now, it only remains to select which of the functions is $latex v$ and which is $latex u$. For example, if you choose:

• $latex u =x$
• $latex du =dx$
• $latex dv =\cos x dx$
• $latex v =\int \cos x dx = \sin x$

Then:

$$\int x\cos x dx= x\sin x-\int \sin xdx =x\sin x+\cos x + C$$

$$\int x\cos x dx=x\sin x+\cos x + C$$

In general, it is advisable to take $latex dv$ as the part of the integrand that is most easily integrated, and $latex u$ as the one that is easiest to derive. The practice facilitates the decision, having as a goal that $latex vdu$ is easy to compute.

### EXAMPLE 2

Calculate the integral:

$$\int xe^x dx$$

Selecting:

• $latex u =x$
• $latex du =dx$
• $latex dv =e^x dx$
• $latex v =\int e^x dx = e^x$

With the formula:

$$\int udv=uv -\int vdu$$

We have:

$$\int xe^x dx=xe^x-\int e^xdx$$

$$=xe^x-e^x+C$$

This result can be factored and we get:

$$\int xe^x dx=e^x(x-1)+C$$

### EXAMPLE 3

Solve the following:

$$\int x \sqrt{x-1}dx$$

In this case, we specify:

• $latex u =x$
• $latex du =dx$
• $latex dv = \sqrt{x-1}\:dx$
• $latex v =\int \sqrt{x-1}\:dx =\dfrac{2}{3} (x-1)^\frac{3}{2}$

Using the formula:

$$\int udv=uv -\int vdu$$

We get:

$$\int x \sqrt{x-1}dx=x\cdot\dfrac{2}{3} (x-1)^\frac{3}{2}-\int \dfrac{2}{3} (x-1)^\frac{3}{2}dx$$

The integral obtained in the last step is now solved:

$$\int \dfrac{2}{3} (x-1)^\frac{3}{2}dx=\dfrac{2}{3}\int (x-1)^\frac{3}{2}dx$$

By means of a simple variable change:

• $latex dw=dx$
• $latex w=x-1$

We have:

$$\int (x-1)^\frac{3}{2}dx=\int w^\frac{3}{2}dw=\left[\dfrac{w^{\frac{3}{2}+1}}{\frac{3}{2}+1}\right]$$

$$=\dfrac{2}{5}w^{\frac{5}{2}}+C=\dfrac{2}{5}(x-1)^{\frac{5}{2}}+C$$

Finally, this result is substituted here:

$$\int x \sqrt{x-1}dx=x\cdot\dfrac{2}{3} (x-1)^\frac{3}{2}-\dfrac{2}{3}\int (x-1)^\frac{2}{3}dx$$

$$=\dfrac{2x}{3} (x-1)^\frac{3}{2}-\dfrac{2}{3}\cdot\dfrac{2}{5}(x-1)^{\frac{5}{2}}+C$$

$$=\dfrac{2x}{3} (x-1)^\frac{3}{2}-\dfrac{4}{15}(x-1)^{\frac{5}{2}}+C$$

$$=2 (x-1)^\frac{3}{2}\left[\dfrac{x}{3} -\dfrac{2}{15}(x-1)\right]+C$$

$$=2 (x-1)^\frac{3}{2}\left[\dfrac{x}{3} -\dfrac{2x}{15}+\dfrac{1}{15}\right]+C$$

$$=2 (x-1)^\frac{3}{2}\left[\dfrac{3x}{15} +\dfrac{2}{15}\right]+C$$

$$=\frac{2}{15}(x-1)^\frac{3}{2}\left(3x +2\right)+C$$

Then, the integral sought is:

$$\int x \sqrt{x-1}dx=\frac{2}{15}(x-1)^\frac{3}{2}\left(3x +2\right)+C$$

## Integration by parts – Practice problems

Integration by parts quiz
You have completed the quiz!

Interested in learning more about integrals? You can take a look at these pages:

### Jefferson Huera Guzman

Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.