Integration by Parts

Undoing the Product Rule

The product rule of differential calculus shows how to differentiate the product of two functions (here called $u,v$): $$\frac{d}{dx}u(x)v(x)=u(x)v^\prime(x)+v(x)u^\prime(x)$$ To turn this rule into a useful tool for anti-differentiation, we take the indefinite integral of both sides: $$u(x)v(x)=\int u(x)v^\prime(x)dx+\int v(x)u^\prime(x)dx$$ At face value this may not appear to be very useful, we have just written one function ($uv$) as a sum of two other integrals! But subtracting one of these integrals to the other side reveals a new perspective: $$\int u(x)v^\prime(x)dx=u(x)v(x)-\int v(x)u^\prime(x)dx$$ Now we can view this as a way of exchanging one integral for another: if you are presented with a problem to antidifferentiate the product $u(x)v^\prime(x)$ (the left hand side), you can succeed if you can 1) calculate $u(x)v(x)$ (that is, find the antiderivative of $v^\prime$), and 2) calculate $\int v(x)u^\prime(x)dx$. In certain situations, this second integral is much simpler than the original, making this a powerful technique to replace one integral with another.

Making a Memorable Notation

If we abbreviate the combination $v^\prime(x)dx$ as simply $dv$ (as we did in $u$-substitution), then the entire integrand becomes $u dv$. Following this convention through and writing $u^\prime(x)dx$ as $du$, we can express the entire formula compactly as

$$\int u dv=uv-\int vdu$$

Choosing $u$ and $v$

To implement this, we need to divide an integrand into the product of two functions; one of which we are going to differentiate (called $u)$ and one of which we will antidifferentiate (named $dv$, to signal that we want its antiderivative, $v$). Main goals: $u$ should get simpler when we differentiate, and we should know how to antidifferentiate $dv$. Here’s some examples to attempt: choose your $u$ and $dv$ following this guideline, then compute $du$ and $v$, and substitute into the formula defining integration by parts:

$$\int x\sin(x) dx$$

$$\int t e^{2t}dt$$

$$\int \frac{\ln(x)}{x^2}dx$$

When $dv=dx$ is the right choice

Sometimes its possible to successfully employ integration by parts by setting the entire integrand to $u$, which just leaves $dx$ to be $dv$. This makes the calculation of $v$ easy (as $v=x$…), and thus converts the original problem $\int u dx$ into $$\int u dx = xu(x)-\int xu^\prime(x)dx$$ So, this technique will succeed if you can manage to integrate $xu^\prime(x)dx$.

$$\int \ln(x)dx$$

$$\int \arctan(x)dx$$ (This one leads to a new integral requiring u-sub!)

Iterated Integration by Parts

Replacing an integral containing $u$ and $dv$ with an integral containing $du$ and $v$ hopefully simplifies the problem, even if this simplification may not directly lead to an indefinite integral that we already know. But thats ok! So long as we managed to make it one step simpler, we can always apply integration by parts again (and again…) to the remaining integral until the problem has completely dissolved! Here’s some examples requiring at least two iterations of integration by parts.

$$\int t^2e^t dt$$

$$\int x^3\sin(x) dx$$

$$\int x^4\sin(2x)dx$$

Interesting Examples

Sometimes in performing integration by parts, s

$$\int e^x\sin x dx$$

$$\int\sin(nx)\cos(mx)dx$$

$$\int \ln\sqrt{x} dx$$

$$\int x\ln(x)dx$$

$$\int \sqrt{x}\ln(x)dx$$

Substitutions that lead to Integration by Parts

$$\int e^{\sqrt{x}}dx$$

$$\int\cos(\ln(x))dx$$

$$\int x\ln(1+x)dx$$