<s>[INST] Improper integral $\int_0^{\infty} \frac{1}{(x+1)(\pi^2+\ln^2(x))}dx$ I am having trouble solving

$$\int_0^{\infty} \frac{1}{(x+1)(\pi^2+\ln^2(x))}dx$$

let $\ln(x)=u$ then $x du= dx $
$$\Rightarrow \int_0^{\infty} \frac{1}{(x+1)(\pi^2+\ln^2(x))}dx=\int_0^{\infty} \frac{x+1-1}{(x+1)(\pi^2+u^2)}du=\int_0^{\infty} \frac{1}{\pi^2+u^2}du-\int_0^{\infty} \frac{1}{(e^u+1)(\pi^2+u^2)}du$$
$$=\frac{1}{2}-\int_0^{\infty} \frac{1}{(e^u+1)(\pi^2+u^2)}du$$
How do you proceed from here?
I put it into an integral calculator and it stated "Antiderivative or integral could not be found."
How do you solve this ?
 [/INST] Enforcing the substitution $x\mapsto e^{-x}$ reveals
$$\int_0^\infty \frac1{(x+1)(\pi^2+\log^2(x))}\,dx=\int_{-\infty}^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx$$
Then, note that
$$\begin{align}
\int_{-\infty}^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx&=\int_{-\infty}^0 \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx+ \int_0^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx\\\\
&=\int_0^\infty \frac{1}{(1+e^{x})(\pi^2+x^2)}\,dx+ \int_0^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx\\\\
&=\int_0^\infty \frac{1}{\pi^2+x^2}\,dx\\\\
&=\frac12
\end{align}$$
</s>