Find $\int \frac {1} {(x-a)^n} dx$

Question

Find $\int \frac {1} {(x-a)^n} dx$ where $n \in \mathbb{N}, a \in \mathbb{R}$

Am I supposed to solve this using substitution?

Answer 1

Yes. Make the $u$-substitution $u = x-a$, which has $du=dx$. Then

$$\int \frac{1}{(x-a)^n} dx \implies \int \frac{1}{u^n}du = \int u^{-n}du$$

Take care when using the power rule here: we have cases for the power rule when $n=1$ and when $n\neq 1$. If $n=1$, then

$$\int u^{-n}du = \int\frac{du}{u} = \ln|u| + C = \ln|x-a|+C$$

If $n\neq 1$, however,

$$\int u^{-n}du = \frac{u^{-n+1}}{-n+1} + C = \frac{(x-a)^{-n+1}}{-n+1} + C$$

Answer 2

If $n=1$ it is $\ln(x-a)$ otherwise it is ${1\over{-n+1}}(x-a)^{-n+1}$.