Evaluating $\int_0^1\frac{x^{-a}-x^{a}}{1-x}\,\mathrm dx$ How to evaluate following integral

$$\int_0^1\frac{x^{-a}-x^{a}}{1-x}\,\mathrm dx$$

I tried the feynman way 
$$\begin{align} 
I'(a)&=\int_0^1\frac{-x^{-a}\ln x-x^{a}\ln x}{1-x}\,\mathrm dx\\
&= \int_0^1\ln x\left(\frac{-x^{-a}-x^{a}}{1-x}\right)\,\mathrm dx\\
\end{align}$$
I don't have any idea about how to proceed!
Some help is appreciated

We can do it without using The Feynman Way
$$\begin{align}
\int_0^1 \frac{x^{-a}-x^a}{1-x}\,\mathrm dx
&=\int_0^1 \left[\frac{x^{-a}}{1-x}-\frac{-x^a}{1-x}\right]\,\mathrm dx\tag{1}\\
&=\int_0^1\sum_{k=1}^\infty\left(x^{k-a-1}-x^{k+a-1}\right)\,\mathrm dx\tag{2}\\
&=\sum_{k=1}^\infty\int_0^1\left(x^{k-a-1}-x^{k+a-1}\right)\,\mathrm dx\tag{3}\\
&= \sum_{n=1}^\infty \left(\frac{1}{k-a} - \frac{1}{k+a}\right)\tag{4}\\
&= \sum_{n=1}^\infty \left(\frac{k+a-k+a}{(k-a)(k+a)}\right)\tag{5}\\
&= \sum_{n=1}^\infty \left(\frac{2a}{k^2-a^2}\right)\tag{6}\\
&= 2a\sum_{n=1}^\infty\frac{1}{k^2-a^2}\tag{7}\\
&= 2a\left[\frac{1}{2a} \left( \frac{1}{a} \, - \, \pi \cot(a\pi) \right)\right]\tag{8}\\
&= \frac{1}{a} \, - \, \pi \cot(a\pi)\tag{9}\\
\end{align}$$
Note that for convergence $a \in (-1,1)$

$$\large\int_0^1 \frac{x^{-a}-x^a}{1-x}\,\mathrm dx= \frac{1}{a} \, - \, \pi \cot(a\pi)$$

