Integration by parts is the inverse of the derivative product rule. It is very useful when $u$-substitution and using standard integration techniques aren’t enough to handle the problem. Integration by parts states that if you have an integral of the form $\displaystyle \int u \: dv$, we may rewrite it as:

\[\begin{align*} \int u \: dv = uv – \int v \: du \end{align*}\]

When using integration by parts, it is very important to pick the correct $u$ and $dv$. In general, when making the selection, we are looking for $u$ to be something that is easy to take the derivative, while $dv$ should be something easy to double integrate. The selection process is very important as picking the wrong $u$ and $dv$ could cause integration by parts to be more harmful than helpful.

Example 1 – Using Integration by Parts

Evaluate

\[\begin{align*} \int x\ln x \: dx \end{align*}\]

Letting $u = \ln x$ and $dv = x \: dx$, we get that $du = \dfrac1x \: dx$ and $v = \dfrac{x^2}{2}$. We may then use integration by parts:

\[\begin{align*} \int x\ln x \: dx &= \ln x \cdot \dfrac{x^2}{2} – \int \dfrac{x^2}{2} \cdot \dfrac1x \: dx\\ &= \dfrac{x^2 \ln x}{2} – \dfrac{1}{2} \int x \: dx\\ &= \boxed{\dfrac{x^2 \ln x}{2} – \dfrac{x^2}{4} + C} \end{align*}\]

The selection of $u$ and $dv$ was easy in this case as working with a polynomial as the integrand is very easy using power rule, and the derivative of $\ln x$ is a common derivative. Flipping our $u$ and $dv$ though tells a different story as $\displaystyle \int \ln x \: dx = x\ln x – x + C$, which returns us to our original problem plus more if we had picked it in the reverse order.

Example 2 – Double Integration by Parts

Evaluate

\[\begin{align*} \int x^2 e^{3x} \: dx \end{align*}\]

For our selection of $u$ and $dv$ this time around, it is less obvious as both are relatively easy to work with. In cases like these, it is generally a better idea to select $u$ as the function that becomes simpler when we differentiate so that we are not stuck in an infinite loop. With that in mind, lets make the selection $u = x^2$ and $dv = e^{3x}$.

\[\begin{align*} \int x^2 e^{3x} \: dx &= x^2 \cdot \dfrac13 e^{3x} – \int \dfrac13 e^{3x} \cdot 2x \: dx \end{align*}\]

The second part of our integration by parts is a product, so once again we can apply integration by parts, letting our new $u = \dfrac23 x$ and $dv = e^3x$ again.

\[\begin{align*} \int x^2 e^{3x} \: dx &= x^2 \cdot \dfrac13 e^{3x} – \left[\dfrac23x \cdot \dfrac13e^{3x} – \int \dfrac13 e^{3x} \cdot \dfrac23 \: dx\right]\\ &= \dfrac13x^2e^{3x} – \dfrac29xe^{3x} + \dfrac29 \int e^{3x} \: dx\\ &= \boxed{\dfrac13x^2e^{3x} – \dfrac29xe^{3x} + \dfrac2{27}e^{3x} + C} \end{align*}\]

Notice that if we let $dv = x^2$ instead at the beginning, we would have had:

\[\begin{align*} \int x^2 e^{3x} \: dx &= e^{3x} \dfrac{x^3}{3} – \int \dfrac{x^3}{3} \cdot 3e^{3x} \: dx \end{align*}\]

We would be stuck in an infinite loop of using integration by parts with this selection as our integrand power has increased to $3$ instead of decreasing like our previous selection of $u$ and $dv$.

Example 3 – Definite Integral \& Trig Example

Evaluate

\[\begin{align*} \int_0^{\pi} x^2\cos(x) \: dx \end{align*}\]

The derivative and integral of both $x^2$ and $\cos x$ are not too difficult. We will select $u = x^2$ so that the power gets smaller instead of larger. With that said, $dv = \cos x$. Using the integration by parts formula:

\[\begin{align*} \int_0^{\pi} x^2\cos(x) \: dx &= x^2 \sin x – \int_0^{\pi} 2x\sin x \: dx\\ &= x^2 \sin x – 2 \int_0^{\pi} x \sin x \: dx \end{align*}\]

Notice that we have a product in our second integral, so we may apply integration by parts a second time. Note, if we had selected our $u$ and $dv$ in the reverse order, our new integral would have $x^3$ instead of $x$, so it was important to let $u = x^2$ and not the other way around. Proceeding this time with $u = x$ and $dv = \sin x$:

\[\begin{align*} &= x^2 \sin x – 2 \int_0^{\pi} x \sin x \: dx\\ &= x^2 \sin x – 2 (-x\cos x – \int_0^{\pi} -\cos x \: dx)\\ &= \biggr[x^2 \sin x + 2x\cos x – 2\sin x\biggr]_0^{\pi}\\ \end{align*}\]

We evaluate the bounds like normal, plugging in $x = \pi$ and subtracting the result of when we plug in $x = 0$. Our final answer is:

\[\begin{align*} \boxed{\int_0^{\pi} x^2\cos(x) \: dx = -2\pi} \end{align*}\]