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: A substitution of $u=x+1$ makes the process much easier using the binomial expansion on the top. 
A: Note: I was trying to do something else but it didn't work out like I thought it would, but i thought I would ost it anyways.
My answer isn't too useful but it is interesting. My answer uses a little differentiation under the integral sign. My answer is a complicated way of doing it your way...
But it was fun!
If anything is unclear in my answer, please ask and Me or someone else will answer.
$ \begin{array}{l}
\int\limits ^{1}_{0}\frac{x^{a-1}}{x+1} dx=\int\limits ^{1}_{0}\frac{x^{a} x^{-1}}{x+1} dx=f( a) \Longrightarrow \\
f'( a) =\int\limits ^{1}_{0}\frac{ln( x) x^{a} x^{-1}}{x+1} dx,\ x=e^{-t} ,\ dx=-e^{-t} dt,\ t( 0) =\infty ,\ t( 1) =0\Longrightarrow \\
\int\limits ^{1}_{0}\frac{ln( x) x^{a} x^{-1}}{x+1} dx=-\int\limits ^{\infty }_{0}\frac{te^{-ta}}{1-\left( -e^{-t}\right)} dt=-\int\limits ^{\infty }_{0} te^{-ta}\sum ^{\infty }_{n=1}\left[\left( -e^{-t}\right)^{n-1}\right] dt=-\sum ^{\infty }_{n=1}\left[( -1)^{n-1}\int\limits ^{\infty }_{0} te^{-ta} e^{t( 1-n)} dt\right]\\
=\sum ^{\infty }_{n=1}\left[( -1)^{n}\int\limits ^{\infty }_{0} te^{-t( n-1+a)} dt\right] ,\ n\in \mathbb{N} ,\ \Re ( a)  >0\Longrightarrow n-1+a\ is positive\\
\therefore \lim _{t\rightarrow \infty }\left( e^{-t( n-1+a)}\right) =0\Longrightarrow \\
\\
\sum ^{\infty }_{n=1}\left[( -1)^{n}\int\limits ^{\infty }_{0} te^{-t( n-1+a)} dt\right] =\sum ^{\infty }_{n=1}\left[( -1)^{n}\left(\lim\limits ^{ \begin{array}{l}
t=\infty \\
\end{array}}_{t=0}\left(\frac{te^{-t( n-1+a)}}{n-1+a}\right) -\frac{1}{1-n-a}\int\limits ^{\infty }_{0} e^{-t( n-1+a)} dt\right)\right]\\
=\sum ^{\infty }_{n=1}\left[( -1)^{n}\left(\frac{1}{( 1-n-a)^{2}}\right)\right]\\
\\
f'( a) =\sum ^{\infty }_{n=1}\left[\frac{( -1)^{n}}{( 1-n-a)^{2}}\right]\\
\therefore f( a) =\sum ^{\infty }_{n=1}\left[( -1)^{n}\int \frac{1}{( 1-n-a)^{2}} da\right]\\
\\
\int \frac{1}{( 1-n-a)^{2}} da,\ t=1-n-a,\ a=1-n-t\Longrightarrow da=-1dt\Longrightarrow \\
\int \frac{1}{( 1-n-a)^{2}} da=-\int t^{-2} da=\frac{1}{t} =\frac{1}{( 1-n-a)} +C\\
\therefore f( a) =\sum ^{\infty }_{n=1}\left[\frac{( -1)^{n}}{( 1-n-a)}\right] =\sum ^{\infty }_{n=0}\left[\frac{( -1)^{n-1}}{-n-t}\right] =\sum ^{\infty }_{n=0}\left[\frac{( -1)^{n}}{n+t}\right] =\Phi ( -1,\ 1,\ a) \Longrightarrow \\
\\
\int\limits ^{1}_{0}\frac{x^{a-1}}{x+1} dx=\Phi ( -1,\ 1,\ a)\\
\\
in\ case\ you\ don't\ know:\\
\\
\Phi ( z,\ s,\ \alpha ) =\sum ^{\infty }_{n=0}\left[\frac{z^{n}}{( n+\alpha )^{s}}\right]\\
\\
\frac{a}{1-r} =\sum ^{\infty }_{n=1} ar^{n-1} ,\ | r| < 1,\ 
\end{array}$
