Is there a nice easy way to evaluate either
$$\int \frac{x^m}{(x - a)^n}dx$$
or
$$\int \frac{(x - b)^m}{(x - a)^n}dx$$
Where $m, n$ are integers? For the first the best I can come up with is turning $x^m$ into $x^m - a^m + a^m$, factoring & continually reducing things, & the second I'd expand x - b & use the first method, but these can get messy!
In fact expanding the binomial expansion is nearly the only approach for evaluating $\int\dfrac{x^m}{(x-a)^n}dx$ and $\int\dfrac{(x-b)^m}{(x-a)^n}dx$ .
For $\int\dfrac{x^m}{(x-a)^n}dx$ ,
When $m$ is a positive integer,
$\int\dfrac{x^m}{(x-a)^n}dx=\int\dfrac{(x-a+a)^m}{(x-a)^n}dx=\int\dfrac{\sum\limits_{k=0}^mC_k^ma^{m-k}(x-a)^k}{(x-a)^n}dx=\int\sum\limits_{k=0}^mC_k^ma^{m-k}(x-a)^{k-n}~dx$
Then integrate it termwisely. Beware whether the $\ln(x-a)$ term exist.
When $n$ is a negative integer,
$\int\dfrac{x^m}{(x-a)^n}dx=\int x^m(x-a)^{-n}~dx=\int x^m\sum\limits_{k=0}^{-n}C_k^{-n}(-1)^ka^kx^{-n-k}~dx=\int\sum\limits_{k=0}^{-n}C_k^{-n}(-1)^ka^kx^{m-n-k}~dx$
Then integrate it termwisely. Beware whether the $\ln x$ term exist.
Similar on $\int\dfrac{(x-b)^m}{(x-a)^n}dx$ .