Evaluate the following integral : $I=\int_{0}^{1}\frac{x^{a-1}}{1+x}\ dx, a>0$? Here is my attempt:
$$\begin{split}\int_{0}^{1}\frac{x^{a-1}}{x+1}\ dx&=\sum_{k=0}^{\infty }(-1)^k\int_{0}^{1}\ x ^{a+k-1}dx\\
&=\sum_{k=0}^{\infty }\frac{(-1)^k}{a+k}\\
&=\sum_{k=0}^{\infty }\left [ \frac{1}{2k+a} -\frac{1}{2k+1+a}\right ]\\
&=\frac{1}{2}\sum_{k=1}^{\infty }\left [ \frac{1}{k+\frac{a-2}{2}}-\frac{1}{k+\frac{a-1}{2}} \right ]\\
&=\frac{1}{2}\left [ \Psi \left (1+\frac{a-1}{2} \right ) -\Psi \left (1+\frac{a-2}{2} \right )\right ]\\
&=\frac{1}{2}\left [ \Psi \left (\frac{a+1}{2} \right )-\Psi \left (\frac{a}{2} \right ) \right ]
\end{split}$$
Can we compute the integral above in different methods?
All approaches are highly appreciated.

A substitution of $u=x+1$ makes the process much easier using the binomial expansion on the top. 
