Evaluation of $ I = \int_{0}^{1}\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx$ 
If $\displaystyle I = \int_{0}^{1}\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx\;,$ Then value of $100(I-\ln 2) =$

$\bf{My\; Try::}$ Let $\cot^{-1}(x)=t\;,$ Then $\displaystyle \frac{1}{1+x^2}dx = -dt$ and changing limit, We get
$$\displaystyle I = -\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\frac{2\cot t-\csc^4 t\cdot t}{1-\csc^2 t\cdot t}dt = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(\frac{2\cot t-\csc^4 t\cdot t}{1-\csc^2 t\cdot t}\right)dt$$
Now How can I solve after that, Help Required
Thanks 
 A: HINT:
$$I=\int\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx =\int\dfrac{\dfrac{2x}{(1+x^2)^2}-\cot^{-1}x}{\dfrac1{1+x^2}-\cot^{-1}(x)}dx$$
As $\dfrac{d\left(\dfrac1{1+x^2}-\cot^{-1}(x)\right)}{dx}=-\dfrac{2x}{(1+x^2)^2}+\dfrac1{1+x^2},$ write
$$I=\int\dfrac{\dfrac{2x}{(1+x^2)^2}-\dfrac1{1+x^2}}{\dfrac1{1+x^2}-\cot^{-1}(x)}dx+\int dx$$
Set $\dfrac1{1+x^2}-\cot^{-1}(x)=u$
A: You can start by rewriting your integrand as
$$
\begin{aligned}
1&+\frac{2x-(1+x^2)}{(1+x^2)\bigl[1-(1+x^2)\,\text{arccot}\,x\bigr]}\\
&=1+\frac{2x\bigl[1-(1+x^2)\,\text{arccot}\,x\bigr]+(1+x^2)\bigl[2x\,\text{arccot}\,x-1\bigr]}{(1+x^2)\bigl[1-(1+x^2)\,\text{arccot}\,x\bigr]}\\
&=1+\frac{2x}{1+x^2}+\frac{1-2x\,\text{arccot}\,x}{1-(1+x^2)\,\text{arccot}\,x}
\end{aligned}
$$
Now, we are really lucky! Noting that $D(1-(1+x^2)\,\text{arccot}\,x)=1-2x\,\text{arccot}\,x$, we find that 
$$
\int \frac{2x-(1+x^2)^2 \,\text{arccot}\,x}{(1+x^2)[1-(1+x^2)\,\text{arccot}\,x]}\,dx=x+\log|1+x^2|+\log|1-(1+x^2)\,\text{arccot}\,x|+C.
$$
I leave the rest for you.
