Seeking to evaluate: $\int_{0}^{1}\arctan(x)\cdot\ln\frac{x+x^3}{(1-x)^2}\cdot\frac{\mathrm dx}{x}$ We wish to evaluate this integral,
$$\int_{0}^{1}\arctan(x)\cdot\ln\frac{x+x^3}{(1-x)^2}\cdot\frac{\mathrm dx}{x}$$
I have tried using substitution, $u=\frac{x+x^3}{(1-x)^2}$ and integration by parts but, I am not able to simplify it down.
What can kind of method can we use to evaluate this integral? 

$$I=\int_0^1\frac{\arctan x}{x}\ln\left(\frac{x+x^3}{(1-x)^2}\right)dx\\=\int_0^1\frac{\arctan x\ln x}{x}dx+\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx$$
For the first integral, write the taylor series of $\arctan x$
$$\int_0^1\frac{\arctan x\ln x}{x}dx=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln x\ dx=-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}=-\beta(3)=-\frac{\pi^3}{32}$$
where $\beta(3)$ is the Dirichlet beta function.
as for the second one, it is evaluated here
$$\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$$
$$\Longrightarrow I=\frac{\pi^3}{32}$$
