Integration By Parts

If you think of $u$-substitution as the reverse chain rule, you can think of this as the reverse product rule

For our purposes, we only know two integration techniques: $u$-sub and integration by parts. Because of this, you can recognize when you must use one or the other

When to use it: you will use integration by parts when you see the product of two functions in the integrand. Unlike $u$-sub, the two functions don't have to share any sort of special relationship.

How to use it:

Start off by labeling one of the functions $u$ and the other $dv$. These are just labels.

Your integral should now look like this: $\displaystyle\int{u}dv$.

The formula for integration by parts, without getting into the proof, is

$$\displaystyle\int u dv=uv-\displaystyle\int v du$$

Set up a table. On the left you will have your $u$, on the right, $dv$. So far it will look like this:

\begin{array}{c|c}
{} & {} \\
\hline
u & dv \\
\end{array}

Now, since you have $u$ and $dv$, and since you need $du$ and $v$ for your equation to work, you need to differentiate the left side of your table ($u$), and integrate the right ($dv$). Your table should now read:

\begin{array}{c|c}
\textrm{Differentiate} & \textrm{Integrate} \\
\hline
u & dv \\
du & v \\
\end{array}

Now you have everything you need for your equation. You simply multiply $u$ by $dv$ and subtract it from the integral of $v\cdot du$. On the table, this can be represented by drawing lines. If we take diagonal lines to represent multiplication (plus to represent the positive product), and horizontal lines to mean finding the integral of the product of the two linked terms (minus to represent subtraction), we can draw the following as the solution to our integral:

And you're done! The expression in the table is now equivalent to $uv-\displaystyle\int v du$.

Nuances

The most common thing people struggle with is understanding how to choose $u$ and $dv$.

Although you can mostly choose **either** of the two functions to be one or the other, there is always a definitive easier choice for both terms.

Think about it: you know that by using this method, you will always be deriving the $u$ term, and always integrating the $dv$ term. Thus, it logically follows that the $dv$ value shouldn't be impossible hard to integrate, and the $u$ value should become simpler when deriving.

In fact, there exists a very handy acronym as a **suggestion** for how to choose $u$ and $dv$. However, this is merely a guide and some integrals may call for different approaches. This acronym is

L ogarithmic functions

I nverse trigonometric functions

A lgebraic (polynomial) functions

T rigonometric functions

E xponential functions

I nverse trigonometric functions

A lgebraic (polynomial) functions

T rigonometric functions

E xponential functions

Following LIATE, we choose $u$ in descending order starting from L (i.e. log functions are preferable to inverse trig functions) and $dv$ in ascending order starting from E (i.e. exponential is preferred over trig).

Tabular Integration by Parts

You may be wondering why we even bothered to express the formula for integration by parts with that table. The reason for this is because, as the name implies, tables are essential to another form of integration by parts, tabular integration.

At times, you will come across a problem which will require you to perform integration by parts more than once. An example of this is $\displaystyle\int x^5e^xdx$. If you were to integrate this using the method described above, you'd be at work for a while! Luckily, there is a shortcut that we can take, and this shortcut is called *tabular integration*.

When to use it: Usually, tabular integration can be applied when the integrand consists of the product of an algebraic function and some other function whose integral follows a pattern (such as $\sin{x}, e^x$, etc).

How to use it: Let's use the example of $\displaystyle\int x^5e^xdx$. By following LIATE, we choose $u=x^5$ and $dv=e^x dx$. We know that the integral of $e^x$ is $e^x$. Since $x^5$ is a polynomial function, we also know that if we derive it enough times, we will eventually get zero. Thus, instead of doing integration by parts over and over, we can set up a table which follows the same rules as our last ones. The **rule of thumb** to keep in mind when doing tabular integration is to interchange the plus/minus signs with every increment. We will derive the left column and integrate the right to get the following:

Remember, diagonal lines indicate multiplication, and the positive and negative signs tell us whether to add or subtract the product (the first line has a positive sign because its positive). Horizontal lines imply integrating, so when we take the integral of 0, we get our constant of integration $+C$. Thus, using our shortcut, we can evaluate the integral to the true, lengthy answer: $\displaystyle\int x^5e^xdx=x^5e^x-5x^4e^x+20x^3e^x-60x^2e^x+120xe^x-120e^x+C$.