Evaluation of Integral $\int_{0}^1 \frac{\arctan x }{1+x} dx$ How do you compute $$\int_{0}^1 \frac{\arctan x }{1+x} dx$$
 A: Using integration by parts 
$$\int_0^1 \frac{\arctan x}{1+x} dx  = \arctan(x) \ln(1+x)|_0^1 - \int_0^1 \frac{\ln (1+x)}{1+x^2}dx$$
The former part is $\displaystyle \frac{\pi}{
4} \ln 2 $ and the latter part is $\displaystyle \frac{\pi}{8} \ln  2$ which is answered here.
A: $$
\begin{align}
\int_0^1\frac{\arctan(x)}{1+x}\,\mathrm{d}x
&=\int_0^{\pi/4}\frac{\theta}{1+\tan(\theta)}\,\sec^2(\theta)\,\mathrm{d}\theta\tag{1}\\[6pt]
&=\int_0^{\pi/4}\frac{\theta\,\mathrm{d}\theta}{\cos(\theta)\,(\cos(\theta)+\sin(\theta))}\tag{2}\\[6pt]
&=\int_0^{\pi/4}\frac{(\frac\pi4-\theta)\,\mathrm{d}\theta}{\cos(\theta)\,(\cos(\theta)+\sin(\theta))}\tag{3}\\[6pt]
&=\frac\pi8\int_0^{\pi/4}\frac{\mathrm{d}\theta}{\cos(\theta)\,(\cos(\theta)+\sin(\theta))}\tag{4}\\[6pt]
&=\frac\pi8\int_0^{\pi/4}\frac{\sec^2(\theta)}{1+\tan(\theta)}\,\mathrm{d}\theta\tag{5}\\[6pt]
&=\frac\pi8\int_0^1\frac1{1+x}\mathrm{d}x\tag{6}\\[6pt]
&=\frac\pi8\Big[\log(1+x)\Big]_0^1\tag{7}\\[12pt]
&=\frac\pi8\log(2)\tag{8}
\end{align}
$$
$(1):$ $x=\tan(\theta)$
$(3):$ $\theta\mapsto\frac\pi4-\theta$
$(4):$ since $(2)=(3)$ we have $(3)=\frac{(2)+(3)}{2}$
$(6):$ $x=\tan(\theta)$
