Yesterday i searched on internet hard integrals, the Question is one of them.i forgot the source(book) of the question
\begin{align}
I &= \int_{0}^{1}\frac{\arctan^2(x)\log(1-x)}{1+x^2}\,dx
\end{align}
\begin{align} u &= \arctan^2(x) \quad \Rightarrow \quad du = 2\arctan(x)\frac{1}{1+x^2}dx \ dv &= \frac{\log(1-x)}{1+x^2}dx \quad \Rightarrow \quad v = \int \frac{\log(1-x)}{1+x^2}dx \end{align}
Now integration by parts
\begin{align} I &= \arctan^2(x)\int \frac{\log(1-x)}{1+x^2}dx – \int \left(2\arctan(x)\frac{1}{1+x^2}\right)\left(\int \frac{\log(1-x)}{1+x^2}dx\right)dx \ &= \arctan^2(x)v – 2\int \frac{\arctan(x)}{1+x^2} \left(\int \frac{\log(1-x)}{1+x^2}dx\right)dx \end{align}
Best Answer
firstly put $x \to \tan x$ $$ I=\int_0^1 \frac{\arctan^2{x}\log(1-x)}{1+x^2} dx=\int_0^{\frac{\pi}{4}} x^2\log(1-\tan x) dx$$ using the formula $$ 1-\tan x=\sqrt{2}\sin\left(\frac{\pi}{4}-x \right)\sec x$$ So $$ I=\int_0^{\frac{\pi}{4}} x^2 \log\left(\sqrt{2}\sin\left(\frac{\pi}{4}-x \right)\sec x \right) dx $$ and we have Fourier series for $\log \sin x $ and $\log \cos x $ proved here $$ \log \sin x =-\log2-\sum_{k=1}^{\infty} \frac{cos(2kx)}{k} \quad,\quad \log \cos x =-\log2-\sum_{k=1}^{\infty} (-1)^k\frac{cos(2kx)}{k}$$ then $$ I=\frac{\log2}{2}\left(\frac{x^3}{3}\right)_0^{\frac{\pi}{4}}+\int_0^{\frac{\pi}{4}} x^2 \sum_{k=1}^{\infty} \frac{1}{k}\left((-1)^k\cos(2kx)-\cos \left(2k\left(\frac{\pi}{4}-x \right)\right) \right) dx$$ $$ =\frac{\log2}{384} \pi^3+\sum_{k=1}^{\infty} \frac{1}{k}\left((-1)^k \int_0^{\frac{\pi}{4}} x^2 \cos(2kx)dx-\int_0^{\frac{\pi}{4}} x^2 \cos \left(2k\left(\frac{\pi}{4}-x \right)\right) dx \right) $$ now you can get this results easily $$ \int_0^{\frac{\pi}{4}} x^2 \cos(2kx)dx=\left(\frac{\pi^2}{32k}-\frac{1}{4k^3} \right) \sin \left(\frac{\pi k}{2} \right)+\frac{\pi}{8k^2} \cos \left(\frac{\pi k}{2} \right) $$ $$ \int_0^{\frac{\pi}{4}} x^2 \cos \left(2k\left(\frac{\pi}{4}-x \right)\right) dx=\frac{\pi}{8k^2}-\frac{1}{4k^3}\sin \left(\frac{\pi k}{2} \right) $$ now we need to evaluate two series let's denote it by $A_1,A_2$ $$A_1=\sum_{k=1}^{\infty} \frac{(-1)^k }{k} \left(\left(\frac{\pi^2}{32k}-\frac{1}{4k^3} \right) \sin \left(\frac{\pi k}{2} \right)+\frac{\pi}{8k^2} \cos \left(\frac{\pi k}{2} \right) \right) $$ $$A_2=\sum_{k=1}^{\infty} \frac{1}{k}\left(\frac{\pi}{8k^2}-\frac{1}{4k^3}\sin \left(\frac{\pi k}{2} \right) \right)=\frac{\pi}{8} \zeta{(3)}-\frac{1}{4}\sum_{k=1}^{\infty} \frac{1}{k^4}\sin \left(\frac{\pi k}{2} \right) $$ using this formula $$ \sum_{k=1}^{\infty} (-1)^k f(k)=\sum_{k=1}^{\infty} \left(f(2k)-f(2k-1) \right)$$ and get $$A_1=\sum_{k=1}^{\infty} \left(\frac{1}{2k} \frac{\pi}{8(2k)^2} (-1)^k-\frac{1}{2k-1} \left(\frac{\pi^2}{32(2k-1)}-\frac{1}{4(2k-1)^3} \right)(-1)^{k-1} \right) $$ $$ =-\frac{\pi}{64} \eta{(3)}-\frac{\pi^2}{32}\beta(2)+\frac{1}{4} \beta(4)=-\frac{3\pi}{256}\zeta(3)-\frac{\pi^2}{32}G+\frac{1}{4} \beta(4) $$ and $$A_2=\frac{\pi}{8} \zeta{(3)}-\frac{1}{4}\sum_{k=1}^{\infty} \frac{1}{(2k-1)^4}(-1)^{k-1}=\frac{\pi}{8} \zeta{(3)}-\frac{\beta(4)}{4} $$ finally $$ I=\frac{\log2}{384} \pi^3+A_1-A_2=\frac{\log2}{384} \pi^3-\frac{35\pi}{256}\zeta(3)-\frac{\pi^2}{32}G+\frac{1}{2} \beta(4)$$