<s>[INST] Evaluate the integral $I=\int_{0}^{\infty}\frac{\ln^3{x}}{(1+x^2)(1+x)^2}dx$ Find this integral
$$I=\int_{0}^{\infty}\dfrac{\ln^3{x}}{(1+x^2)(1+x)^2}dx$$
My try: let $x=\tan{t}$
then
$$I=\int_{0}^{\frac{\pi}{2}}\dfrac{\ln^3{\tan{t}}}{(1+\tan{t})^2}dt$$
I am unable to simplify after this. This problem is from QQ.
 [/INST] Here is an elementary way. First note that
$$\int_1^{\infty} \dfrac{\ln^3(x) dx}{(1+x^2)(1+x)^2} = \int_1^0 \dfrac{-\ln^3(x)}{(1+1/x^2)(1+1/x)^2} \dfrac{-dx}{x^2} = \int_0^1 \dfrac{-x^2 \ln^3(x)}{(1+x^2)(1+x)^2}dx$$
Hence, your integral is
$$I = \int_0^1 \dfrac{(1-x^2) \ln^3(x)}{(1+x^2)(1+x)^2}dx = \underbrace{\int_0^1 \dfrac{\ln^3(x)}{1+x}dx}_J - \overbrace{\int_0^1 \dfrac{x\ln^3(x)}{1+x^2}dx}^K$$
We have
$$J = \sum_{k=0}^{\infty}(-1)^k\int_0^1 x^k \ln^3(x)dx = \sum_{k=0}^{\infty}(-1)^{k+1} \dfrac6{(k+1)^4} = -\dfrac7{120} \pi^4 \tag{$\star$}$$
We have
$$K = \sum_{k=0}^{\infty}(-1)^k\int_0^1 x^{2k+1} \ln^3(x)dx = \sum_{k=0}^{\infty}(-1)^{k+1} \dfrac3{8(k+1)^4} = -\dfrac7{1920} \pi^4 \tag{$\dagger$}$$
Hence,
$$\boxed{\color{red}{I = J-K = -\dfrac7{128}\pi^4}}$$
Where we used the following facts
$$\sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(k+1)^4} = -\dfrac78 \zeta(4)$$
$$\zeta(4) = \dfrac{\pi^4}{90}$$
to simplify $(\star)$ and $(\dagger)$.
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