<s>[INST] Evaluation of $\int_{0}^{1}\frac{\arctan x}{1+x}dx$ 
Evaluation of $$\int_{0}^{1}\frac{\tan^{-1}(x)}{1+x}dx = \int_{0}^{1}\frac{\arctan x}{1+x}dx$$

$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{\tan^{-1}(ax)}{1+x}dx$$
Then $$\frac{dI}{da} = \frac{d}{da}\left[\int_{0}^{1}\frac{\tan^{-1}(ax)}{1+x}\right]dx = \int_{0}^{1}\frac{x}{(1+a^2x^2)(1+x)}dx$$
So we get $$I = \frac{1}{a^2}\int_{0}^{1}\frac{x}{(x+1)(x^2+k^2)}dx\;,$$ Where $\displaystyle k=\frac{1}{a^2}$
Now Using Partial fraction for $$\frac{x}{(x^2+k^2)(x+1)} = \frac{Ax+B}{x^2+k^2}+\frac{C}{x+1} = \frac{(A+C)x^2+(A+B)x+(B+Ck^2)}{(x^2+k^2)(x+1)}$$
So we get $A+C=0$ and $A+B=1$ and $B+Ck^2=0$
So we get $\displaystyle A=\frac{1}{1+k^2}$ and $\displaystyle B = \frac{k^2}{1+k^2}$ and $\displaystyle C=-\frac{1}{1+k^2}$
So $$\frac{dI}{da} = \frac{1}{a^2(1+k^2)}\int_{0}^{1}\left[\frac{x+k^2}{x^2+k^2}-\frac{1}{x+1}\right]dx$$
So $$\frac{dI}{da} = \frac{1}{a^2(1+k^2)}\left[\frac{1}{2}\ln(x^2+k^2)-k\tan^{-1}\left(\frac{x}{k}\right)-\ln(x+1)\right]_{0}^{1}$$
So we get $$\frac{dI}{da} = \frac{1}{a^2(1+k^2)}\left[\frac{1}{2}\ln(1+k^2)-\ln(k)-k\tan^{-1}\left(\frac{1}{k}\right)-\ln 2\right]$$
So So we get $$\frac{dI}{da}=\frac{1}{1+a^2}\left[\frac{1}{2}\ln(1+a^2)-\frac{1}{a}\tan^{-1}(a)-\ln2\right]$$
Now integration of above expression is very lengthy, Is there is any other method
If yes Then plz explain here, Thanks
 [/INST] The substitution $x\mapsto \frac{1-x}{1+x}$ transforms the integral into
$$I=\int_0^1 \frac{\tan^{-1} x}{1+x}\,dx=\int_0^1 \frac{\frac{\pi}{4}-\tan^{-1} x}{1+x}\,dx.$$
Because $\displaystyle\,\,\,\, \tan^{-1} \frac{1-x}{1+x}=\tan^{-1}1-\tan^{-1} x=\frac{\pi}{4}-\tan^{-1} x.$ 
Taking the average of these two representations then gives
$$I=\frac12\int_0^1 \frac{\frac{\pi}{4}}{1+x}dx=\frac{\pi}{8}\ln2.$$
</s>