Evaluation of $\sum_{n=1}^{\infty} (\frac{1}{4n-3}-\frac{1}{4n-1}) $ Prove that 
$$\sum_{n=1}^{\infty} (\frac{1}{4n-3}-\frac{1}{4n-1})=\frac{\pi}{4}$$
Could someone give me slight hint as how to convert above expression to $\int_{0}^{1} \tan^{-1}x.dx$

By IBP,
\begin{eqnarray*}\int \tan^{-1}x.dx&=&x\times tan^{-1}x|-\int x\times \frac{1}{1+x^2}.dx \\
&=&x\times tan^{-1}x|-\int \frac{d(x^2+1)}{2dx}\times \frac{1}{1+x^2}.dx\\
&=&x\times tan^{-1}x|-\ln|x^2+1| |
  \end{eqnarray*}
