Integration by parts

The process of finding antiderivatives can feel a bit frightening as integrands become more complicated. Luckily there are certain techniques that can help us identify and adequately tackle more intricate indefinite integrals. In this topic, you'll take a look at the Integration by Parts technique. You'll also see how it can help you when integrands involve the product of two (or more) functions.

Integration by Parts

As its name suggests, this method consists of separating the original integrand into parts. Before elaborating further, let's recall the rule to find derivatives of products of two functions $u(x)$ and $v(x)$

$$(u \cdot v)' = u' \cdot v + u \cdot v'$$Rearranging terms

$$u \cdot v' = (u \cdot v)' - u' \cdot v$$Integrating both sides of the equation

$$\int u \cdot v' \ dx = u \cdot v - \int u' \cdot v \ dx$$We know that $du = u'(x) \ dx$ and $dv = v'(x) \ dx$, so our expression becomes

$$\boxed{\quad \int u \cdot dv = u \cdot v - \int v \cdot du \quad}$$which is the main formula for this method.

Here, $u \cdot dv$ represents the original integrand, including the differential. This integrand represents the product of two (or more) functions: $u$ is usually a function that becomes easier to integrate after differentiation. $dv$ is the product of $dx$ times a function whose antiderivative $v$ is known beforehand or can be determined without much effort.

If applied properly, the resulting integral $\int v \cdot du$ should be at least easier to solve than the original one. It can even be immediately solved as in one of the cases you've seen previously.

For example, let's consider the following integral

$$\int x e^x dx$$First, let's define both $u$ and $dv$ as follows:

$$u = x \qquad dv = e^x dx$$Then, you need to find both $v$ and $du$:

$$\frac{du}{dx} = 1 \implies du = dx \qquad \int dv=\int e^x dx \implies v=e^x$$Now, you can substitute these expressions following the formula above, like so

$$\int u \cdot dv = u \cdot v - \int v \cdot du \implies \int x \cdot e^x dx = x \cdot e^x - \int e^x \cdot dx$$You can see how your initial integral has become a simpler one. In fact, you know the resulting integral has a direct antiderivative:

$$\int e^x \cdot dx = e^x + C$$Thus,

$$\begin{align*} \int x \cdot e^x dx &= x \cdot e^x - e^x + C \\ &= e^x \cdot (x-1) + C \end{align*}$$

Integral of logarithmic functions

In order to find the antiderivative of a logarithmic function

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

you can use the integration by parts method. By rewriting your integrand, you can identify both $u$ and $v$ and their derivatives:

$$\int \ln(x)dx \ = \int \ln(x) \cdot 1 \ dx \implies \begin{cases} u =\ln(x) &\implies du = \frac{1}{x} dx \\ dv = 1 \cdot dx &\implies v = x \end{cases}$$Substituting these expressions in your integration by parts formula, you obtain

$$\int \ln(x) \cdot 1 \ dx = \ln(x) \cdot x - \int \frac{1}{x} \cdot x \ dx$$And so,

$$\begin{align*} \int \ln(x)dx &= x \ln(x) - \int \frac{1}{x} \cdot x \ dx \\ &= x \ln(x) - \int 1 \ dx \\ &= x \ln(x) - x + C \\ \int \ln(x)dx &= x \cdot (\ln(x) - 1) + C \end{align*}$$

You can generalize this to logarithms of any base using the change of base property of logarithms:

$$\log_ab = \frac{\log_cb}{\log_ca}$$Then,

$$\begin{align*} \int \log_a(x)dx = \frac{1}{\ln(a)} \int \ln(x)dx &= x \cdot \left(\frac{\ln(x)-1}{\ln(a)}\right) + C \\ \int \log_a(x)dx &= x \cdot \left(\log_a(x) - \frac{1}{\ln(a)}\right) + C \end{align*}$$

Algebra and integrals

Let's consider the following integral

$$I = \int \sin(x) \cdot 2^x \ dx$$

Where $I$ is a placeholder for the result of the integral.

You can apply integration by parts in order to expand the expression, like so

$$\int 2^x \cdot \sin(x) \ dx \implies \begin{cases} u =2^x &\implies du = \ln(2) \cdot 2^x \ dx \\ dv = \sin(x)dx &\implies v = -\cos(x) \end{cases}$$Thus,

$$I = -\cos(x) \cdot 2^x + \int \cos(x) \cdot 2^x \cdot \ln(2) \ dx = -\cos(x) \cdot 2^x + \ln(2) \int \cos(x) \cdot 2^x \ dx$$Now, you have a new integral to solve. Applying integration by parts again

$$\int 2^x \cdot \cos(x) \ dx \implies \begin{cases} u =2^x &\implies du = \ln(2) \cdot 2^x \ dx \\ dv = \cos(x)dx &\implies v = \sin(x) \end{cases}$$Then,

$$\begin{align*} I &= -\cos(x) \cdot 2^x + \ln(2) \int \cos(x) \cdot 2^x \ dx \\ &= -\cos(x) \cdot 2^x + \ln(2) \cdot \left[ \sin(x) \cdot 2^x - \int \sin(x) \cdot 2^x \cdot \ln(2) \ dx \right] \\ I&= -\cos(x) \cdot 2^x + \ln(2) \cdot \sin(x) \cdot 2^x - \ln^2(2) \int \sin(x) \cdot 2^x \ dx \end{align*}$$But wait! What is in that last term on the right-hand side of your equation? Recall your original integral:

$$I = \int \sin(x) \cdot 2^x \ dx$$Substituting in, and solving for $I$

$$\begin{align*} I&= -\cos(x) \cdot 2^x + \ln(2) \cdot \sin(x) \cdot 2^x - \ln^2(2) \cdot I \\ (\ln^2(2) + 1) \cdot I&= -\cos(x) \cdot 2^x + \ln(2) \cdot \sin(x) \cdot 2^x \\ I&= \frac{-\cos(x) \cdot 2^x + \ln(2) \cdot \sin(x) \cdot 2^x}{\ln^2(2) + 1} + C\\ \end{align*}$$[ALERT] Remember to add the constant of integration! [/ALERT]

Finally, you have

$$\int \sin(x) \cdot 2^x \ dx = \frac{-\cos(x) \cdot 2^x + \ln(2) \cdot \sin(x) \cdot 2^x}{\ln^2(2) + 1} + C$$

For some integrands, the cyclical relationship between their derivatives and antiderivatives appears when using the integration-by-parts method. As you saw in this example, which deals with an integrand comprised of sines and cosines, the re-appearing integrand can be grouped together with the original one to solve these integrals.

Tips for newcomers

The actual skill behind the Integration by Parts method lies in developing an intuition to correctly identify both $u$ and $dv$. This intuition is built with practice by solving integrals over and over, and there's no way around that. However, when it comes to new learners trying to solve their first integrals, there are a few useful tips to remember:

• Derivatives of logarithmic, inverse trigonometric, and polynomial functions all result in "simpler" expressions: $$\begin{align*} & \frac{d}{dx} \ln(x) = \frac{1}{x} \\ \\ & \frac{d}{dx} \arctan(x) = \frac{1}{1+x^2} \\ \\ &\frac{d}{dx} (x^2 + 5x + 3) = x + 5 \\ \end{align*}$$While their antiderivatives are more intricate:$$\begin{align*} & \int \ln(x)dx = x \cdot (\ln(x) - 1) + C \\ \\ & \int \arctan(x) \ dx = x \cdot \arctan(x) - \frac{1}{2} \ln |x^2 + 1| + C \\ \\ & \int (x^2 + 5x + 3) \ dx = \frac{x^3}{3} + \frac{5x^2}{2} + 3x +C \\ \end{align*}$$Thus, these types of functions are good candidates for $u$

• Neither differentiation nor integration of exponential and (non-inverse) trigonometric functions usually result in "simpler" integrals: $$\begin{align*} & \frac{d}{dx} e^x = e^{x} \\ \\ & \frac{d}{dx} \sin(x) = \cos(x) \\ \\ & \int e^x \ dx = e^{x} + C \\ \\ & \int \sin(x) \ dx = - \cos(x) + C \\ \end{align*}$$If differentiating another factor $u$ in the original integrand simplifies the whole expression, it can be beneficial to integrate these. Thus, these types of functions are good candidates for $dv$

An example:

Let's consider the following integral:

$$\int x \sin(x) \ dx$$Identifying both $u$ and $dv$ is a pretty straightforward process using our tips:

$$\int x \cdot \sin(x) \ dx \implies \begin{cases} u =x &\implies du = dx \\ dv = \sin(x)dx &\implies v = -\cos(x) \end{cases}$$Then, you are left with an easier integral to solve:
$$\int x \sin(x) \ dx = -x \cos(x) + \int \cos(x) dx$$

A counterexample:

It is good to remember that these tips are not rules, but guidelines to which there can be many exceptions. As a counter-example, let's consider the following integral:

$$\int \frac{x e^x}{(x + 1)^2} \ dx$$If you were to choose $u$ and $dv$ as follows

$$\int \frac{x e^x}{(x + 1)^2} \ dx \implies \begin{cases} u = \frac{x}{(x+1)^2} &\implies du = \frac{x-1}{(x+1)^3} dx \\ dv = e^x dx &\implies v = e^x \end{cases}$$You would be left with an even more difficult integral to solve

$$\int \frac{x e^x}{(x + 1)^2} \ dx = \frac{x e^x}{(x+1)^2} - \int \frac{(x-1) \ e^x}{(x+1)^3} dx$$If instead, you were to choose $u$ and $dv$ in this other manner:

$$\int \frac{x e^x}{(x + 1)^2} \ dx \implies \begin{cases} u = xe^x &\implies du = e^x \cdot(x+1) \ dx \\ dv = \frac{1}{(x+1)^2} dx &\implies v = - \frac{1}{x+1} \end{cases}$$You would obtain a much easier integral to solve

$$\int \frac{x e^x}{(x + 1)^2} \ dx = -\frac{x e^x}{x+1} + \int e^x dx$$In summary, a good rule of thumb is comparing the resulting integral to the original one; if the integrand became more difficult to integrate then it probably wasn't the right choice. Remember, there is no shame in going back and choosing again!

Conclusion

In this topic, you have learned about the integration by parts method. This integration method can be summarized as follows:

$$\int u \cdot dv = u \cdot v - \int v \cdot du$$

In addition, you've also learned the rule to determine the antiderivative for logarithmic functions:

$$f(x) = \log_a(x) \implies \int \log_a(x)dx = x \cdot \left(\log_a(x) - \frac{1}{\ln(a)}\right) + C$$

You also learned that certain integrands can re-appear after applying the integration by parts method and that elementary algebra can help us solve these integrals.

Finally, you learned that even though certain functions can usually fit better as either $u$ or $dv$, there can be exceptions. Therefore, an intuition must be developed in order to make sure the resulting integral is easier to solve that the original one.