Integration-by-Parts Formula

Learning Outcomes

• Recognize when to use integration by parts.
• Use the integration-by-parts formula to solve integration problems.
• Use the integration-by-parts formula for definite integrals.

If, $h\left(x\right)=f\left(x\right)g\left(x\right)$, then by using the product rule, we obtain ${h}^{\prime }\left(x\right)={f}^{\prime }\left(x\right)g\left(x\right)+{g}^{\prime }\left(x\right)f\left(x\right)$. Although at first it may seem counterproductive, let’s now integrate both sides of this equation: $\displaystyle\int {h}^{\prime }\left(x\right)dx=\displaystyle\int \left(g\left(x\right){f}^{\prime }\left(x\right)+f\left(x\right){g}^{\prime }\left(x\right)\right)dx$.

This gives us

$h\left(x\right)=f\left(x\right)g\left(x\right)=\displaystyle\int g\left(x\right){f}^{\prime }\left(x\right)dx+\displaystyle\int f\left(x\right){g}^{\prime }\left(x\right)dx$.

Now we solve for $\displaystyle\int f\left(x\right){g}^{\prime }\left(x\right)dx:$

$\displaystyle\int f\left(x\right){g}^{\prime }\left(x\right)dx=f\left(x\right)g\left(x\right)-\displaystyle\int g\left(x\right){f}^{\prime }\left(x\right)dx$.

By making the substitutions $u=f\left(x\right)$ and $v=g\left(x\right)$, which in turn make $du={f}^{\prime }\left(x\right)dx$ and $dv={g}^{\prime }\left(x\right)dx$, we have the more compact form

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

Integration by Parts

Let $u=f\left(x\right)$ and $v=g\left(x\right)$ be functions with continuous derivatives. Then, the integration-by-parts formula for the integral involving these two functions is:

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

The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. The following example illustrates its use.

Example: Using Integration by Parts

Use integration by parts with $u=x$ and $dv=\sin{x}dx$ to evaluate $\displaystyle\int x\sin{x}dx$.

Try It

Evaluate ${\displaystyle\int }^{\text{ }}x{e}^{2x}dx$ using the integration-by-parts formula with $u=x$ and $dv={e}^{2x}dx$.

Watch the following video to see the worked solution to the above Try It

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

The natural question to ask at this point is: How do we know how to choose $u$ and $dv?$ Sometimes it is a matter of trial and error; however, the acronym LIATE can often help to take some of the guesswork out of our choices. This acronym stands for Logarithmic Functions, Inverse Trigonometric Functions, Algebraic Functions, Trigonometric Functions, and Exponential Functions. This mnemonic serves as an aid in determining an appropriate choice for $u$.

The type of function in the integral that appears first in the list should be our first choice of $u$. For example, if an integral contains a logarithmic function and an algebraic function, we should choose $u$ to be the logarithmic function, because L comes before A in LIATE. The integral in the previous example has a trigonometric function $\text{(}\sin{x}\text{)}$ and an algebraic function $\left(x\right)$. Because A comes before T in LIATE, we chose $u$ to be the algebraic function. When we have chosen $u$, $dv$ is selected to be the remaining part of the function to be integrated, together with $dx$.

Why does this mnemonic work? Remember that whatever we pick to be $dv$ must be something we can integrate. Since we do not have integration formulas that allow us to integrate simple logarithmic functions and inverse trigonometric functions, it makes sense that they should not be chosen as values for $dv$. Consequently, they should be at the head of the list as choices for $u$. Thus, we put LI at the beginning of the mnemonic. (We could just as easily have started with IL, since these two types of functions won’t appear together in an integration-by-parts problem.) The exponential and trigonometric functions are at the end of our list because they are fairly easy to integrate and make good choices for $dv$. Thus, we have TE at the end of our mnemonic. (We could just as easily have used ET at the end, since when these types of functions appear together it usually doesn’t really matter which one is $u$ and which one is $dv$.) Algebraic functions are generally easy both to integrate and to differentiate, and they come in the middle of the mnemonic.

To use the by-parts technique successfully, it is helpful to first review the derivative rules of several familiar transcendental functions.

Recall: Derivative rules for transcendental functions

1. $\frac{d}{dx} (\sin x) = \cos x$
2. $\frac{d}{dx} (\cos x) = -\sin x$
3. $\frac{d}{dx} (\ln x) = \frac{1}{x}$
4. $\frac{d}{dx} (\arcsin x) = \frac{1}{\sqrt{1-x^2}}$
5. $\frac{d}{dx} (\arctan x) = \frac{1}{1+x^2}$

example: Using integration by parts

Evaluate $\displaystyle\int \frac{\text{ln}x}{{x}^{3}}dx$.

try it

Evaluate ${\displaystyle\int }^{\text{ }}x\text{ln}xdx$.

Watch the following video to see the worked solution to the above Try It

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

In some cases, as in the next two examples, it may be necessary to apply integration by parts more than once.

example: Applying integration by parts more than once

Evaluate ${\displaystyle\int }^{\text{ }}{x}^{2}{e}^{3x}dx$.

Example: Applying Integration by Parts When LIATE Doesn’t Quite Work

Evaluate ${\displaystyle\int }^{\text{ }}{t}^{3}{e}^{{t}^{2}}dt$.

Example: Applying Integration by Parts More Than Once

Evaluate ${\displaystyle\int }^{\text{ }}\sin\left(\text{ln}x\right)dx$.

try it

Evaluate ${\displaystyle\int }^{\text{ }}{x}^{2}\sin{x}dx$.

Watch the following video to see the worked solution to the above Try It

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

Integration by Parts for Definite Integrals

Now that we have used integration by parts successfully to evaluate indefinite integrals, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.

Integration by Parts for Definite Integrals

Let $u=f\left(x\right)$ and $v=g\left(x\right)$ be functions with continuous derivatives on $\left[a,b\right]$. Then

${\displaystyle\int }_{a}^{b}udv={uv|}_{a}^{b}-{\displaystyle\int }_{a}^{b}vdu$.

Example: finding the area of a region

Find the area of the region bounded above by the graph of $y={\tan}^{-1}x$ and below by the $x$ -axis over the interval $\left[0,1\right]$.

Example: Finding a volume of revolution

Find the volume of the solid obtained by revolving the region bounded by the graph of $f\left(x\right)={e}^{\text{-}x}$, the x-axis, the y-axis, and the line $x=1$ about the y-axis.

try it

Evaluate ${\displaystyle\int }_{0}^{\pi \text{/}2}x\cos{x}dx$.

Watch the following video to see the worked solution to the above Try It

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.