How to compute this integral... I need to compute $$ \int_0^1 \frac{\arctan x}{(1+x)^2}dx$$
I tried substituting $t= \arctan x$ (check please), getting $$ \int \frac{t}{1+2\sin t \cos t}dt $$ but I still can't handle it. Any help would be useful: I haven't done integrals for years. Thanks

By parts $f=\arctan x$ and $g=-\frac{1}{x+1}$
$$-\frac{\arctan x}{x+1}+\int\frac{dx}{(x+1)(x^2+1)}$$
Now partial fractions
$$-\frac{\arctan x}{x+1}+\frac 1 2\int \frac{1-x}{x^2+1}dx+\frac 1 2\int\frac{dx}{x+1}\\
=-\frac{\arctan x}{x+1}-\frac 1 2\int \frac{x}{x^2+1}dx+\frac 1 2\int\frac{dx}{x^2+1}+\frac 1 2 \int \frac {dx}{x+1}\\
=\frac{\ln(x+1)}{2}-\frac{\ln(x^2+1)}{4}-\frac{\arctan x}{x+1}+\frac {\arctan x}{2}+\mathcal C$$
$$\left(\frac{\ln(x+1)}{2}-\frac{\ln(x^2+1)}{4}-\frac{\arctan x}{x+1}+\frac {\arctan x}{2}\right)\bigg|_0^1=\frac{\ln(2)}{4}$$
