Find $\int _0^{\infty }\frac{\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x$ In what ways can I evaluate $$\int _0^{\infty }\frac{\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x$$
I tried a few methods but none work or simplify things and i cant think of substitutions that could turn things better, while the substitution $x=1/x$ gets
$$\int _0^{\infty }\frac{\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x-\int _0^{\infty }\frac{x^2\ln \left(x\right)}{1-x^2+x^4}\:\mathrm{d}x$$
i dont know how to proceed, splitting the integral at point 1 also doesnt help much.
The integral can also be expressed as
$$I=\int _0^1\frac{\left(1+x^2\right)\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x-\int _0^1\frac{x^2\ln \left(x\right)}{1-x^2+x^4}\:\mathrm{d}x$$
that first integral as mentioned in the comments has been evaluated and is
$$I=\frac{\pi }{6}\ln \left(2+\sqrt{3}\right)-\int _0^1\frac{x^2\ln \left(x\right)}{1-x^2+x^4}\:\mathrm{d}x$$
But how to tackle the second one?
 A: Split the integration range
\begin{align}
I& =\int _0^{\infty }\frac{\ln (1+x)}{1-x^2+x^4}dx= \int _0^1\frac{(1+x^2)\ln (1+x)-x^2\ln x}{1-x^2+x^4}dx\\
\end{align}
Then, integrate by parts via
$$d\left( \cot^{-1}\frac x{x^2-1}\right)=\frac{1+x^2}{1-x^2+x^4}dx$$
$$d\left( \frac12\tan^{-1}\frac x{1-x^2} - \frac1{2\sqrt3}\tanh^{-1}\frac {\sqrt3x}{1+x^2}\right)= \frac{x^2}{1-x^2+x^4}dx
$$
to express the integral as
\begin{align}
 I= I_1 -\frac1{2\sqrt3}I_2 +\frac12I_3\tag1
\end{align}
where
$$
I_1 = \int_0^1 \frac{dx}{1+x} \cot^{-1}\frac {x}{1-x^2},\>\>\>\>\>
I_2 = \int_0^1 \frac{dx}{x} \tanh^{-1}\frac {\sqrt3x}{1+x^2}\\
I_3 = \int_0^1 \frac{dx}{x} \tan^{-1}\frac {x}{1-x^2}
$$
To evaluate $I_1$, let $J_1(a) =\int_0^1 \frac{dx}{1+x} \cot^{-1}\frac {2x\sin a}{1-x^2}$
\begin{align}
J_1’(a) &= \int_0^1 \frac{2\cos a (x-x^2)dx}{(x^2+1)^2-(2x\cos a)^2}= - \frac\pi4\tan\frac a2+\frac12\left( a\>{\csc a}+ \ln\tan\frac a2\right)
\end{align}
\begin{align}
I_1 &=J_1(\frac\pi6) = J_1(0)+\int_0^{\frac\pi6}J_1’(a)da \\
&= \frac\pi2\ln2-\frac\pi4 \int_0^{\frac\pi6}\tan\frac a2 da+\frac12\int_0^{\frac\pi6} d\left( a\ln\tan\frac a2\right)\\
&=\frac\pi2\ln2 -\frac\pi2\ln\cos\frac\pi{12}-\frac\pi{12}\ln\tan\frac\pi{12}= \frac\pi6\ln(2+\sqrt3)\tag2
\end{align}
To evaluate $I_2$, let $J_2(a) =\int_0^1 \frac{dx}{x} \tanh^{-1}\frac {2ax}{1+x^2}$
\begin{align}
J_2’(a) &= \int_0^1 \frac{2 (1+x^2)}{(x^2+1)^2-(2ax)^2}dx =  \frac\pi2\frac1{\sqrt{1-a^2}}
\end{align}
\begin{align}
I_2 &=J_2(\frac{\sqrt3}2) = \int_0^{\frac{\sqrt3}2}J_2’(a)da = \frac\pi2 \int_0^{\frac{\sqrt3}2} \frac{da}{\sqrt{1-a^2}}=\frac{\pi^2}6\tag3
\end{align}
To evaluate $I_3$
$$I_3 = \int_0^1 \frac{dx}{x} \tan^{-1}\frac {x}{1-x^2}
= \int_0^1 \frac{\tan^{-1}x}x dx+ \int_0^1\underset{x^3\to x}{\frac{\tan^{-1}x^3}x}dx\\
= \frac43\int_0^1 \frac{\tan^{-1}x}x dx= \frac43G\tag4
$$
Plug (2), (3) and (4) into (1) to obtain
$$I=\frac\pi6\ln(2+\sqrt3) -\frac{\pi^2}{12\sqrt3} +\frac23G
$$
