Compute $\int\limits_0^1 \log (2\arctan x+\frac{2x}{1+x^2})(1+x\arctan x)\ dx$ How to Find the value of $$I=\int_{0}^{1}\log \Big(2\arctan x+\frac{2x}{1+x^2}\Big)\Big(1+x\arctan x\Big)dx$$
I tried to subsititution:x=tant, but the process became very complicated,I don't know how to deal with it, and any help will be appreciated.
 A: Well, we have:
$$\mathcal{I}:=\int\limits_0^1\ln\left(2\cdot\arctan\left(x\right)+\frac{2x}{1+x^2}\right)\cdot\left(1+x\cdot\arctan\left(x\right)\right)\space\text{d}x\tag1$$
Using integration by parts:
$$\mathcal{I}=\frac{2+\pi}{4}\cdot\ln\left(\frac{2+\pi}{2}\right)-\int\limits_0^1\frac{1}{1+x^2}\space\text{d}x=\frac{2+\pi}{4}\cdot\ln\left(\frac{2+\pi}{2}\right)-\frac{\pi}{4}\tag2$$

Integration by parts is used as follows:
$$\int\text{f}\left(x\right)\cdot\text{g}\space'\left(x\right)\space\text{d}x=\text{f}\left(x\right)\cdot\text{g}\left(x\right)-\int\text{f}\space'\left(x\right)\cdot\text{g}\left(x\right)\space\text{d}x\tag3$$
Where:

*

*$$\text{f}\left(x\right):=\ln\left(2\cdot\arctan\left(x\right)+\frac{2x}{1+x^2}\right)\tag4$$

*$$\text{g}\space'\left(x\right):=1+x\cdot\arctan\left(x\right)\tag5$$
And then you'll find:

*

*\begin{equation}
\begin{split}
\left[\text{f}\left(x\right)\cdot\text{g}\left(x\right)\right]_0^1&=\lim_{x\space\to\space1}\left(\text{f}\left(x\right)\cdot\text{g}\left(x\right)-\text{f}\left(x-1\right)\cdot\text{g}\left(x-1\right)\right)\\
\\
&=\text{f}\left(1\right)\cdot\text{g}\left(1\right)-\lim_{x\space\to\space0}\text{f}\left(x\right)\cdot\text{g}\left(x\right)\\
\\
&=\frac{2+\pi}{4}\cdot\ln\left(\frac{2+\pi}{2}\right)
\end{split}\tag6
\end{equation}


*$$\text{f}\space'\left(x\right)\cdot\text{g}\left(x\right)=\frac{1}{1+x^2}\tag7$$
