Partial integration

Sometimes, it’s not possible to evaluate the integral just by using the properties of an integral, or even by substituting some expressions.

When those methods aren’t effective, there is another way: partial integration. This is used when one of the factors of the product can be simplified through integration and the other one can be simplified through differentiation. For example, $$\int \ln x$$ can be solved using partial integration.

Let’s see how!

What is partial integration?

Partial integration — or integration by parts — is a process that helps find the integral of a product of functions using the formula:

where $$u$$ is the part of the product easy to differentiate and $$dv$$ easy to integrate.

However, when the partial integration is done, we still need to use the integral rules to solve it.

Need a little refresher on the rules and properties of integration? Here they are:

Why is partial integration so useful?

As was already mentioned, some integrals are just too stubborn to be solved using properties of integrals or by substitution method. Partial integration is often used when the integrand is a product.

Pro tip: When you see $$\ln x$$ in the integrand, you should always use partial integration, and the natural logarithm should always be the factor that is differentiated!

How to use partial integration

Ready to work through some example problems together? Let’s get to it!

Find the integral:

Is your mental alarm sounding? The integrand contains $$\ln x$$, so we need to use partial integration! We know that this integral is not a table integral, and we can’t see any expression that can be substituted — but we do know that the derivation of $$\ln x$$ is $$\frac{1}{x}$$. So, we need to find the function that can be easily differentiated. Remember that by multiplying any expression with one, the expression doesn’t change, so multiply the integrand by $$1$$:

$$\int \ln x\times 1dx$$

Now that we have a product (and we know that $$\ln x$$ can be easily differentiated and $$1$$ can be easily integrated), let $$u=\ln x$$ and $$dv=1dx$$. To use the partial integration formula, we’ll first need to determine $$du$$ and $$v$$.

$$\int \ln x\times 1dx=\left|\begin{array}{cc}u=\ln x& du=?\\ dv=1dx& v=?\end{array}\right|$$

To determine $$du$$ and $$v$$, find the differential using $$du=u^{\prime }dx$$ and integrate $$dv$$:

$$\int \ln x\times 1dx=\left|\begin{array}{cc}u=\ln x& du=(\ln x{)}^{\prime }dx\\ dv=1dx& v=\int 1dx\end{array}\right|$$

As we mentioned, the derivative of $$\ln x$$ is $$\frac{1}{x}$$, and the integral $$\int 1dx$$ is equal to $$x$$, so use substitution:

$$\int \ln x\times 1dx=\left|\begin{array}{cc}u=\ln x& du=\frac{1}{x}dx\\ dv=1dx& v=x\end{array}\right|$$

We’ve determined everything we need for the partial integration, so remember the partial integration formula $$\int udv=uv-\int vdu$$ and substitute the corresponding elements:

$$\int \ln x\times 1dx=\ln x\times x-\int x\times \frac{1}{x}dx$$

The integrand on the right-hand side of the equation can be simplified by canceling the common factor $$x$$, so let’s do that:

$$\int \ln x\times 1dx=\ln x\times x-\int 1dx$$

Remember: the integral $$\int 1dx$$ is equal to $$x$$:

$$\int \ln x\times 1dx=\ln x\times x-x$$

Since the derivative of a constant is zero, we need to add the constant of integration $$C$$ to include all possible anti-derivative functions in the result:

$$\int \ln x\times 1dx=\ln x\times x-x+C,C\in \mathbb{R}$$

We did it! The indefinite integral $$\int \ln xdx$$ is equal to:

Find the integral:

Okay, we know this is not a table integral, nor is there any expression that can be substituted. We also know that the integrand is a product. So, we need to find the function that can be simplified when differentiated, and the other function that can be easily integrated. Since the derivative of $$x$$ is equal to $$1$$, we can conveniently take $$u=x$$ and $$dv=\sin xdx$$.

$$\int x\sin (x)dx=\left|\begin{array}{cc}u=x& du=?\\ dv=\sin xdx& v=?\end{array}\right|$$

To determine $$du$$ and $$v$$, find the differential using $$du=u^{\prime }dx$$ and integrate $$dv$$:

$$\int x\sin (x)dx=\left|\begin{array}{cc}u=x& du=(x{)}^{\prime }dx\\ dv=\sin xdx& v=\int \sin xdx\end{array}\right|$$

As we mentioned, the derivative of $$x$$ is $$1$$ and the integral $$\int \sin xdx$$ is equal to $$-\cos x$$, so use substitution:

$$\int x\sin (x)dx=\left|\begin{array}{cc}u=x& du=1dx\\ dv=\sin xdx& v=-\cos x\end{array}\right|$$

Since we’ve got all that we need for partial integration, keep in mind the partial integration formula $$\int udv=uv-\int vdu$$ and substitute the corresponding elements:

$$\int x\sin (x)dx=x\times (-\cos x)-\int (-\cos x)\times 1dx$$

Notice that the integrand on the right-hand side of the equation can be simplified by using the property $$\int a\times f(x)dx=a\times \int f(x)dx$$:

$$\int x\sin (x)dx=x\times (-\cos x)+\int \cos x\times 1dx$$

Remember that, when multiplying the expression with $$1$$, the result is the expression itself:

$$\int x\sin (x)dx=x\times (-\cos x)+\int \cos xdx$$

Remember that the integral $$\int \cos xdx$$ is equal to $$\sin x$$ and simplify the first product on the right-hand side:

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

Since the derivative of a constant is zero, we need to add the constant of integration $$C$$ to include all possible anti-derivative functions in the result:

$$\int x\sin (x)dx=-x\times \cos x+\sin x+C,C\in \mathbb{R}$$

So the indefinite integral $$\int x\sin (x)dx$$ is equal to:

That wasn’t so bad, right? Now that we’ve walked through a few detailed examples, let’s review the overall process so you use it whenever you need:

Do it yourself!

Whether you’re feeling confident or confused after those examples, it wouldn’t hurt to work on a few practice problems!

Find the integral:

1. $$\int x\times \cos (x)dx$$
2. $$\int (2x+1)e^{x}dx$$
3. $$\int x^2\ln (x)dx$$
4. $$\int e^{x}xdx$$

Solutions:

1. $$x\times \sin x+\cos x+C,C\in \mathbb{R}$$
2. $$2x{e}^x-e^x+C,C\in \mathbb{R}$$
3. $$\frac{\ln x\times x^3}{3}-\frac{x^3}{9}+C,C\in \mathbb{R}$$
4. $$x{e}^x-e^x+C,C\in \mathbb{R}$$

Getting stuck? Scan the problem using your Photomath app, and we’ll walk you through step-by-step!

Here’s a sneak peek of what you’ll see: