# Integration by Parts of Indefinite Integrals

We are now going to learn another method apart from U-Substitution in order to integrate functions. First recall the product rule for a function in the form $f(x) = u(x)v(x)$:

(1)Now suppose that we integrate both sides of the product rule. By the The Fundamental Theorem of Calculus Part 1, it follows that:

(2)Now since $\frac{dv}{dx} = v'(x)$, it follows that $dv = v'(x) \: dx$ and similarly, $du = u'(x) \: dx$, and therefore $\int u(x) \: dv = u(x)v(x) - \int v(x) \: du$, or:

(3)Our general goal with integration by parts is to take a rather difficult integral, $\int u \: dv$ and form a simpler-to-evaluate integral, $\int v \: du$. Thus, in general, we generally want to choose a function $u(x)$ that when differentiated is "simpler".

We will now look at some examples of integration by parts.

## Example 1

**Evaluate the following integral:** $\int xe^x \: dx$.

Let $u = x$, and let $e^x = dv$. We must figure out what our $dx$ is and what our $v$ is. We will summarize this information in the following table:

$u = x$ | $du = dx$ | $v = e^x$ | $dv = e^x$ |

We can now substitute these into our equality $\int u \: dv = uv - \int v \: du$. We thus get:

(4)## Example 2

**Evaluate the following integral:** $\int x \sin 2x \: dx$.

Let $u = x$. It thus follows that $dv = \sin 2x dx$. Hence $du = dx$ and $v = -\frac{1}{2} \cos 2x$. We can thus apply the rule for integration by parts:

(5)## Example 3

**Evaluate the following integral:** $\int e^x \cos x \: dx$.

For this integral, let $u = e^x$ and let $dv = \cos x dx$. It thus follows that $du = e^x \: dx$ and $v = \sin x$. Now let's apply our integration by parts identity.

(6)We're now going to use integration by parts once again. Let's keep $u = e^x$, but now let's introduce $p$ and let $p = \sin x dx$. It follows that $du = e^x \: dx$ and $p = -\cos x$. Making the substitution we obtain:

(7)## Example 4

**Evaluate the following integral:** $\int \tan ^{-1} x \: dx$

Let $u = \tan ^{-1} x$ and $dv = dx$. Therefore $du = \frac{1}{1 + x^2} dx$ and $v = x$. Thus it follows by the integration by parts identity that:

(8)We can now use substitution for the remaining integral. Let $p = 1 + x^2$. Then $dp = 2x \: dx$, so we can make the following substitution:

(9)