<s>[INST] Compute $\int_0^{\pi/4}\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)} x\exp(\frac{x^2-1}{x^2+1}) dx$ 
Compute the following integral
  \begin{equation}
\int_0^{\Large\frac{\pi}{4}}\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\, \exp\left[\frac{x^2-1}{x^2+1}\right]\, dx
\end{equation}

I was given two integral questions by my teacher. I can answer this one although it took a lot of time to compute it. I want to share this problem to the other users here and I would love to see how Mathematics SE users compute this monster. Thank you.
 [/INST] From the numerator, collect the logarithmic terms first. 
$$\displaystyle \int_0^{\pi/4} x\frac{(1+x^2)+(x^2-1)\ln\left(\frac{1-x^2}{1+x^2}\right)}{(1-x^4)(1+x^2)}\exp\left(\frac{x^2-1}{x^2+1}\right)\,dx$$
Rewrite $(1-x^4)=(1-x^2)(1+x^2)$ and divide the numerator by $(1+x^2)$.
$$\displaystyle \int_0^{\pi/4} \frac{x}{(1-x^2)(1+x^2)}\left(1+\frac{x^2-1}{x^2+1}\ln\left(\frac{1-x^2}{1+x^2}\right)\right)\exp\left(\frac{x^2-1}{x^2+1}\right)\,dx $$
Use the substitution $\displaystyle \frac{x^2-1}{x^2+1}=t \Rightarrow \frac{4x}{(1+x^2)^2}\,dx=dt$ to get:
$$\displaystyle \frac{1}{4}\int_{a}^{-1}\frac{e^t}{t}\left(1+t\ln(-t)\right)\,dt= \frac{1}{4}\int_{a}^{-1}e^t\left(\frac{1}{t}+\ln(-t)\right)\,dt$$
where $\displaystyle a=\frac{\pi^2/16-1}{\pi^2/16+1}$
Since $\displaystyle \int e^x(f'(x)+f(x))\,dx=e^xf(x)+C $, the above definite integral is:
$$\displaystyle \frac{1}{4}\left(e^t \ln(-t) \right|_{a}^{-1}=-\frac{1}{4}e^a\ln(-a) \approx \boxed{0.284007} $$ 
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