How to show this integral equality? The given equality is 

$$b \int_0^1 \frac{x^{a-1}}{b+x^n}\,dx=\sum_{k=0}^{\infty}\frac{(-1)^k}{b^k(nk+a)}$$

i tried using Riemann sum but had no luck , any idea how to proceed 
 A: Try manipulating the integrand into a geometric series:
$$ \dfrac{x^{a-1}}{b+x^n} = \frac{x^{a-1}}{b}\dfrac{1}{1+x^n/b} = \frac{x^{a-1}}{b}\sum\limits_{k=0}^\infty(-1)^k\left(\frac{x^n}{b}\right)^k = \frac{1}{b}\sum\limits_{k=0}^\infty(-1)^k\frac{1}{b^k}x^{nk+a-1}  $$
Now integrate, which we can do inside the interval of convergence. 
A: Use the binomial theorem, which states (with some restrictions) the following:
$(a+b)^n = \sum_{k=0}^∞ a^k b^{n-k} {n \choose k}$.
So if the sum converges, your left integrand becomes
$x^{a-1} \sum_{k=0}^∞ (x^n)^k b^{-1-k} {-1 \choose k}$.
In a roundabout way, ${-1 \choose k} = {(-1)}^k$ for a nonnegative integer k.
If all is well-behaved, summation and integration commute, so your left side becomes
$\sum_{k=0}^∞ \int_0^1 x^{a-1+nk} b^{1-1-k} {(-1)}^k dx$,
which simplifies to
$\sum_{k=0}^∞ b^{-k} {(-1)}^k \int_0^1 x^{a-1+nk} dx$.
That leaves you with integrals of the form
$\int_0^1 x^{a-1+nk} dx$,
which are just integrals from zero to one over powers of x, which a small amount of calculus will convert to
$\frac{1}{a+nk}$.
And $b^{-k}$ is just $\frac{1}{b^k}$, so your sum becomes
$\sum_{k=0}^∞ \frac{{(-1)}^k}{(b^k )(a+nk)}$.
And that's the right side. I've glossed over a bunch of ways this might go wrong (your infinite series might not converge sometimes, or b could be zero, to name a few), but I think that in any case in which you're using this identity, you probably already know about those.
