How Do You Do Repeated Integration by Parts?

Repeated integration by parts means applying the integration-by-parts formula more than once. Choose $u$ so that repeated differentiation makes it simpler, then continue until the remaining integral can be evaluated directly.

This is IB Mathematics: Analysis and Approaches HL content from AHL 5.16: Integration techniques. It can be tested analytically on Paper 1 or Paper 2.

The formula is

$$\int u\,\mathrm{d}v=uv-\int v\,\mathrm{d}u.$$

For products involving a polynomial and an exponential or trigonometric function, usually choose the polynomial as $u$. The LIATE guideline supports this choice: logarithmic, inverse trigonometric, algebraic, trigonometric, then exponential.

The repeated process is:

1. Select $u$ using LIATE and choose $\mathrm{d}v$ so that it integrates easily.
2. Apply the formula once, keeping the minus sign visible.
3. Repeat on the new integral. If $u$ is a polynomial, its degree should decrease each time.
4. Substitute backwards, simplify, and add $+C$ once.

Consider

$$I=\int x^2e^x\,\mathrm{d}x.$$

First choose $u=x^2$ and $\mathrm{d}v=e^x\,\mathrm{d}x$. Therefore, $\mathrm{d}u=2$ and :

$$I=x^2e^x-\int 2xe^x\,\mathrm{d}x.$$

Apply integration by parts again:

$$\int 2xe^x\,\mathrm{d}x=2xe^x-\int 2e^x\,\mathrm{d}x =2xe^x-2e^x.$$

Substitute this result:

$$I=x^2e^x-\left(2xe^x-2e^x\right)+C =e^x\left(x^2-2x+2\right)+C.$$

The method terminates because $x^2$ differentiates to $2x$, then $2$, then $0$. Each repetition therefore reduces the algebraic factor by one degree. Differentiate the final expression to check that it returns $x^2e^x$.

For a cyclic integral such as $\int e^x\sin x\,\mathrm{d}x$, repeated integration by parts eventually reproduces the original integral. Move that term to the other side and solve algebraically for the integral.

A common misconception is that each repetition requires switching which factor is $u$. Usually, keep differentiating the polynomial until it becomes zero. Track subtraction signs carefully and include $+C$ only in the final answer.

Exam technique: show the formula or a correct integration-by-parts setup before simplifying. Examiners award method marks for valid choices of $u$, $\mathrm{d}v$, $\mathrm{d}u$, and $v$, even if a later arithmetic error occurs.