Integration by Parts

So far, we've learned about a couple of basic methods of integration: for instance, we know that integrals add: $$\int [f(x)-g(x)]\, dx=\int f(x)\, dx-\int g(x)\, dx.$$

We also get the following integrals:

Integrals of polynomials: $$\int x^n\, dx=\frac{1}{n+1}x^{n+1}+C.$$

Integrals of $\sin{x}$ and $\cos{x}$: $$\int \sin{x}\, dx=-\cos{x}+C.$$ $$\int \cos{x}\, dx=\sin{x}+C.$$

Integrals of exponential functions: $$\int e^x\, dx=e^x+C.$$

So, we can find integrals of functions such as $$\int \left[x+e^x\right]\, dx.$$

As of right now, we don't know enough to figure this out. But that's why we have integration by parts! Specifically, it states the following:

Integration by parts: Suppose $u(x)$ and $v(x)$ are continuously differentiable functions. Then $$\int [u(x)v'(x)]\, dx=u(x)v(x)-\int [u'(x)v(x)]\, dx.$$

We can also write this as $$\int u\, dv=uv-\int v\, du.$$

Do you see how this technique is the counterpart of the product rule for derivatives? First, think about it!

To see why, let's rewrite the equation as $$u(x)v(x)=\int [u(x)v'(x)]\, dx+\int [u'(x)v(x)]\, dx\\=\int [u(x)v'(x)+u'(x)v(x)]\, dx.$$ Aha! This resembles the product rule for derivatives: $$[u(x)v(x)]'=u(x)v'(x)+u'(x)v(x).$$ Let's see what happens when we integrate both sides: $$\int [u(x)v(x)]'\, dx=\int [u(x)v'(x)+u'(x)v(x)]\, dx.$$ By the Fundamental Theorem of Calculus, the left hand side is equal to $u(x)v(x)$. So, we get $$u(x)v(x)=\int u(x)v'(x)\, dx+\int u'(x)v(x)\, dx,$$ which is precisely the formula given above.

Now that we know where this technique comes from, let's try and use this technique for the previous problem! To do so, we let $u(x)=x$ and $dv=e^x\, dx$. Then $$\frac{du}{dx}=1\Longrightarrow du=dx$$ and $$v(x)=\int \, dv=\int e^x\, dx=e^x.$$ (Here, we can ignore the $``+C''$ term.) So, plugging it into the formula for integration by parts $$\int x\cdot e^x\, dx=\int u\, dv=uv-\int v\, du=x\cdot e^x-\int e^x\, dx=x\cdot e^x-e^x+C.$$

In general, the trick is to make $u(x)$ the function that can ``disappear'' when you differentiate it. (For example, in the previous example we set $u(x)=x$ because $u'(x)=1$.

Now you try!

Let's try this together: As before, we want to make $u(x)$ the function whose derivative disappears quickly. So, we set $u(x)=x$ and $dv=\sqrt{x+1}\, dx$. Then $$\frac{du}{dx}=1\Longrightarrow du=dx$$ and $$v(x)=\int\, dv=\int \sqrt{x+1}\, dx=\frac{2}{3}(x+1)^{\frac{3}{2}}.$$ So, plugging these values into the formula for integration by parts, we have $$\int u\, dv=uv-\int v\, du$$ $$\Longrightarrow \int x\sqrt{x+1}\, dx=x\left(\frac{2}{3}(x+1)^{\frac{3}{2}}\right)-\int \frac{2}{3}(x+1)^{\frac{3}{2}}\, dx$$ $$=\frac{2}{3}x(x+1)^{\frac{3}{2}}-\frac{4}{15}(x+1)^{\frac{5}{2}}+C,$$ and we're done.

Let's look at one more problem together:

As you've probably noticed, the integrand appears to have only one function! Well, not quite. We can see that the integral is actually $$\int 1\cdot \ln{x}\, dx.$$ So, one trick would be to let $u=\ln{x}$ and $dv=dx$. So, $$\frac{du}{dx}=\frac{1}{x}\Longrightarrow du=\frac{1}{x}\, dx$$ and $$v(x)=\int \, dv = \int \, dx = x.$$ So, using integration by parts we get $$\int \ln{x}\, dx=\int u\, dv=uv-\int v\, du=x\ln{x}-\int x\cdot \left(\frac{1}{x}\, dx\right)$$ $$=x\ln{x}-\int\, dx=x\ln{x}-x+C.$$ So, integration by parts helped again (this time, a bit sneakily)!

Like in the previous problems, his technique becomes very useful as we approach more complicated integrals. Find out how else this can be used on the next section!

Integration by parts is a powerful tool that results from reversing the Product Rule. Of course, while

$\int{f'(x)g(x) + f(x)g'(x)dx} = f(x)g(x)$

That said, integration by parts results from the formula derived as follows:

$\int{f'(x)g(x)dx} + \int{f(x)g'(x)dx} = f(x)g(x)$

Now subtract the second integral from both sides of the equation:

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

Therefore, at least one of the two functions must have an antiderivative for this to work.

Here is a simple example: integrate $\int{x^2dx}$ using integration by parts.

Obviously $\int{x^2dx} = \frac{1}{3}x^3 + C$ as you're integrating a polynomial, but this example is good for showing how integration by parts works. In this case, $f'(x) = x$ and $g(x) = x$. We must find $f(x)$ and $g'(x)$ at the beginning to ensure integration by parts is usable:

$f(x) = \frac{1}{2}x^2$

$g'(x) = 1$

*leave off the $+C$ until the end. It's not necessary to have multiple constants.

Now plug the two functions and their derivatives into the formula:

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

$\int{x \cdot xdx} = \frac{1}{2}x^2 \cdot x - \int{\frac{1}{2}x^2 \cdot 1dx} \Rightarrow$

$\int{x^2dx} = \frac{1}{2}x^3 - \int{\frac{1}{2}x^2dx} \Rightarrow$

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

Now we can solve for $\int{x^2dx}$ in this equation:

$\frac{3}{2}\int{x^2dx} = \frac{1}{2}x^3 \Rightarrow$

$\int{x^2dx} = \frac{1}{3}x^3$

Then, once everything else is done, you add on the arbitrary constant:

$\int{x^2dx} = \frac{1}{3}x^3 + C$

Although usable for monomials, integration by parts is more commonly used for functions involving exponential, logarithmic, and/or trigonometric functions as factors or terms. Knowledge of some basic antiderivatives is important.

How do we do an integration that is the product of 2 functions? For example... how do we integrate $x \sin3x$ dx?

If $x$ or $\sin3x$ were by themselves we could easily do the integral. But, they are multiplied together and not by themselves, and we don't have a substitution that we can use for this integral. However, we would be able to do this integration by using Integration By Parts.

The formula for Integration By Parts is: $$\int u\, dv=uv-\int v\, du.$$

Supposing $u(x)$ and $v(x)$ are continuously differentiable functions, this formula turns a complicated integral into a much more simpler one. However, it is important to choose u and dv carefully in order to make the integration simple.

So, to integrate $x\sin3x$, we must choose a $u$ and a $dv$. $u$ is usually chosen so that the derivative of $u$ ($\frac{du}{dx}$) turns out to be simpler than $u$.

Let $u=x$ and $dv=\sin3x$

Then $du=dx$ and $v=\frac{-\cos3x}{3}$

Now, we just substitute these 4 expressions into the integration by parts formula to get... $$\int u\, dv=uv-\int v\, du.$$ $$\int x\, \sin3x= x\frac{-\cos3x}{3}-\int\frac{-\cos3x}{3}dx.$$

and simplify to get = $\frac{-x\cos3x}{3} + \frac{\sin3x}{9} + C$

First we will talk about how we derive the formula for integration by parts from the product rule of derivation. The product rule states that $d/dx[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)$.

So we must first integrate both sides

$f(x)g(x)=\int f'(x)g(x)\,dx+\int f(x)g'(x)\,dx$

Then rearrange

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

Now lets try to solve $\int xsinx\,dx$.

In order to use the formula above we must designate values for the terms $f(x), f'(x), g(x)$ and $g'(x)$.

We will start with $f(x)$ and $g'(x)$. What our goal should be is the designate the value of $f(x)$ so that $f'(x)$ is simpler than $f(x)$ and designate $g'(x)$ so that its derivative does not become any more complicated.

This means $f'(x)=1$ which is a simplification from $x$.

Which means $g(x)=-\cos x$

It helps to write all of your values down in one area.

$f(x)=x$, $f'(x)=1$, $g(x)=-\cos x$, $g'(x)=\sin x$

Now we can simply plug into the equation. $\int f(x)g'(x)\,dx=f(x)g(x)-\int f'(x)g(x)\,dx$

$\int x\sin x\,dx=-x\cos x-\int -\cos x\,dx$

And we can evaluate this equation much more easily.

$\int x\sin x\,dx=-x\cos x-(-\sin x)=\sin x-x\cos x$

A very common mistake people make with integration is seen below:

if one were to plug in $f(x)=x$ and $g(x)=1$ you will instantly realize that the idea above that one was going is wrong.

You would get $\int 1x\,dx = \int x\,dx * \int1\,dx$

This further simplifies to $.5x^2 + c = (.5x^2 + c) * (x+c)$ So one can clearly see the addition of a component that makes it so that the two sides cannot be equal thus rendering the above formula wrong.

A better formula to use is "integration by parts" which utilizes substitution and an inverse of the product rule to get the answer for integrals with two separate functions contained within them.

Here we have the formula:

Now lets try using integration by parts to solve the prior integral.

So we have: $\int 1x\,dx$

lets let u = x and dv = 1 so du = 1 and v = x

now lets plug them into the equation

We have: $\int 1x\,dx = x^2 - \int x\,dx$

Which simplifies to $\int 1x\,dx = x^2 - (.5x^2 )+c$

So we know that $\int 1x\,dx$ is really just $\int x\,dx$ and the one is just a place holder for a tougher function for (g(x)). Knowing this, we know that $\int x\,dx = .5x^2 + c$ so going back to the simplification we see that there is a subtraction between $x^2 - .5x^2$ and one realizes that this subtracts to $.5x^2$

So the final equation becomes:

$.5x^2 + c = .5x^2 + c$ which proves that "integration by parts" is valid and works.

Lets look at a problem now. Be careful that you properly choose which function is U and which is dv because you may end up making things harder if you choose the wrong function for each.

Now to properly solve step by step.

$\int x\cos x\,dx = x\sin x - \int \sin x\,dx$

Further solve

$\int x\cos x\,dx = x\sin x - (-\cos x) + C$

Further solve

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

and this is your final answer!

In Calculus 1 we learned about the product rule for differentiation

$\frac{d}{dx} uv = duv+udv$

Integration by parts is the method for reversing the product rule. Reorganizing this formula we get the form

$udv = \frac{d}{dx} uv - duv$

If we see an indefinite or definite integral in the form $\int(udv)$, the form of a function with an antiderivative that used the chain rule, we can use integration by parts.

The formula for integration by parts (IBP) is

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

What this method allows us to do is replace an integral that might be really difficult to take with its derivative.

For example, let's integrate the function $xe^x$; $\int(xe^x)dx$

First we have to make the function look like the IBP form.

Choose which values each variable should stand for and remember that you want to set $u$ equal to whichever value is more difficult to integrate.

Now that our function looks like the form for IBP, we can integrate.

If $u = x$ then $du = dx$

Choosing $x$ as our value for $u$ is a strategic decision because when we take the derivative, $x$ will become 1, making our next step much more simple to integrate; thus is the power of Integration by Parts.

If $dv = (e^x)dx$ then $v = e^x$

Plug these values into the equivalent formula for IBP: $\int(udv) = uv - \int(edu)$

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

Which equals

$xe^x - e^x + C$

and looks like the formula that we began with: $udv = \frac{d}{dx} uv - duv$

(Video) Integration by Parts Made Easy

by PatrickJMT

Introduction

In this video by PatrickJMT, he goes over how to use integration by parts. Integration by parts is, in a way, a more advanced $u$-substitution.

1. You start by selecting one expression to be equal to $u$, and the other expression to be equal to $dv$.
2. You take the derivative of $u$ to get $du$ and take the integral of $dv$ to get $v$.
3. You substitute these into the expression: $uv-\int v\,du$.
4. Solve.

One important part is selecting which expression will be $u$ and which will be $dv$. There is a trick to use. Warning: Teachers will advise you against using this technique, saying that it doesn't always work. While this might be true, I've never encountered a problem where it doesn't work.

We use the term LIATE to determine which expression will be the $u$.

• **$\color{red}{L}$**ogarithms
• **$\color{red}{I}$**nverse Trig
• **$\color{red}{A}$**lgebra
• **$\color{red}{T}$**rig
• **$\color{red}{E}$**xponents

Example 1

It's probably easier to understand all this once we work through a problem. The first problem that Patrick looks at is: $$\int xe^{-x}\,dx$$

Step 1 is to select which expression we want to be $u$ and which to be $dv$. In this problem we have two choices: $x$ and $e^{-x}$. We can use LIATE to help us pick $u$. Algebra, which is $x$, comes before Exponentials, which is $e^{-x}$. So we let $u=x$ and $dv=e^{-x}\,dx$. $$\begin{array}{rl|l} \text{Let:}&u=x&dv=e^{-x}\,dx \end{array}$$

Step 2 is to take the derivative of $u$ and the integral of $dv$. $$\begin{array}{rl|l} \color{red}{\text{Let:}}&\color{red}{u=x}&\color{red}{dv=e^{-x}\,dx}\\ \text{Then:}&du=dx&v=-e^{-x} \end{array}$$

Step 3 is to plug these into the formula: $uv-\int v\,du$ $$\begin{align} uv-\int v\,du&=x(-e^{-x})-\int(-e^{-x})\,dx\\ &=-xe^{-x}+\int e^{-x}\,dx \end{align}$$

Step 4 is to solve. The remaining integral is much easier for us. In fact we literally just took it to get from $dv$ to $v$. So: $$\begin{align} -xe^{-x}+\int e^{-x}\,dx&=-xe^{-x}+(-e^{-x})+C\\ &=-xe^{-x}-e^{-x}+C\\ &=-e^{-x}(x+1)+C \end{align}$$

Example 2

We can use integration by parts to find the integral of $\ln x$. $$\int\ln x\,dx$$

So let's start by choosing $u$. LIATE tells us that $u$ should just be $\ln x$. So what's $dv$? That can just be $dx$! $$\begin{array}{c|c} \color{red}{u=\ln x}&v=x\\ \downarrow&\uparrow\\ du=\frac1x\,dx&\color{red}{dv=dx} \end{array}$$

Now plug these into the formula: $uv-\int v\,du$ and solve. $$\begin{align} uv-\int v\,du&=(\ln x)(x)-\int (x)\left(\frac1x\right)\,dx\\ &=x\ln x-\int dx\\ &=x\ln x-x+C \end{align}$$

Example 3

Let's try one more. $$\int x5^x\,dx$$

LIATE tells us algebra is before exponents, so let's let $u=x$. $$\begin{array}{c|c} \color{red}{u=x}&v=\frac{5^x}{\ln 5}\\ \downarrow&\uparrow\\ du=dx&\color{red}{dv=5^x\,dx} \end{array}$$

Now let's integrate. $$\begin{align} uv-\int v\,du&=\frac{x5^x}{\ln 5}-\int\frac{5^x}{\ln 5}\,dx\\ &=\frac{x5^x}{\ln 5}-\frac{1}{\ln5}\int5^x\,dx\\ &=\frac{x5^x}{\ln 5}-\frac{1}{\ln5}\left(\frac{5^x}{\ln 5}\right)+C\\ &=\frac{x5^x}{\ln 5}-\frac{5^x}{(\ln5)^2}+C\\ &=\frac{5^x}{\ln5}\left(x-\frac{1}{\ln5}\right)+C \end{align}$$

(Video) Integration by Parts with Definite Integrals

by PatrickJMT

Introduction

Here's a video by PatrickJMT in which he shows you how to integrate my parts when the integral has limits of integration. This isn't any more complicated than before, but you just have to be careful. $$\int_a^bu\,dv=uv|_a^b-\int_a^b v\,du$$

It's important to apply the limits of integration to the $uv$ section.

**Example **

The example that Patrick goes over is: $$\int_1^2\frac{\ln x}{x^2}\,dx$$

We start, as always, by picking our $u$ term. LIATE tells us that Logarithm is before Algebra, so $u=\ln x$. $$\begin{array}{c|c} \color{red}{u=\ln x}&v=-\frac1x\\ \downarrow&\uparrow\\ du=\frac1x\,dx&\color{red}{dv=\frac{1}{x^2}\,dx} \end{array}$$ $$\begin{align} uv|_a^b-\int_a^b v\,du&=-\frac{\ln x}{x}\bigg|_1^2-\int_1^2-\frac1{x^2}\,dx\\ &=-\frac{\ln x}{x}\bigg|_1^2+\int_1^2\frac1{x^2}\,dx\\ &=-\frac{\ln x}{x}\bigg|_1^2+-\frac1x\bigg|_1^2\\ &=-\frac{\ln x}{x}\bigg|_1^2-\frac1x\bigg|_1^2\\ &=\left[\frac{\ln2}{2}-\frac{\ln1}{1}\right]-\left[\frac12-\frac11\right]\\ &=\frac{\ln2}{2}+\frac12\\ &=\frac12(\ln2+1) \end{align}$$