# 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\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

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.

## Answer

From the numerator, collect the logarithmic terms first.

Rewrite $(1-x^4)=(1-x^2)(1+x^2)$ and divide the numerator by $(1+x^2)$.

Use the substitution $\displaystyle \frac{x^2-1}{x^2+1}=t \Rightarrow \frac{4x}{(1+x^2)^2}\,dx=dt$ to get:

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:

Attribution
Source : Link , Question Author : Anastasiya-Romanova 秀 , Answer Author : Pranav Arora