Question

Integral of x^2/x^4-1

Answer 1

Answer:

$\displaystyle\int\frac{x^2}{x^4-1}\,dx\\\\=\tfrac12\int\left(\frac1{x^2+1}+\frac1{x^2-1}\right)dx\\\\=\tfrac12\int\frac{dx}{x^2+1}\ +\ \tfrac12\int\frac{dx}{x^2-1}\\\\=\tfrac12\tan^{-1}x + \tfrac14\ln\left|\frac{x-1}{x+1}\right|+C\\\\=\tfrac12\tan^{-1}x - \tfrac12\tanh^{-1}x+C$