TITLE: What is the value of $\int_0^1 \frac{\arctan x}{1+x^{2}} dx$?
QUESTION [7 upvotes]: $$\int_0^1 \frac{\arctan x}{1+x^{2}} dx = ?$$ 
I tried to evaluate it, but I do not know if it is good:
$$\int_0^1\frac{\arctan x}{x^2+1}\,\mathrm{d}x=\left[\arctan x=t\Rightarrow \frac{\mathrm{d}x}{1+x^2}=\mathrm{d}t\right]=\int_0^{\frac{\pi}{4}}t \, \mathrm{d}t=\left[\frac{t^2}{2}\right]_0^{\frac{\pi}{4}}=\frac{\pi^2}{32}$$

REPLY [4 votes]: In general:
$$\int f'(x)f(x)^n\,dx=\frac{f(x)^{n+1}}{n+1}+C$$
So in our case, with
$$f(x)=\arctan x\,\,,\,\,f'(x)=\frac{1}{1+x^2}\Longrightarrow\int\frac{1}{1+x^2}\arctan x\,dx=\frac{1}{2}\arctan^2 x+C$$