Evaluate the following integrals :\(\int\frac{x\,tan^{-1}x^2}{1+x^4}\)

1.	Evaluate the following integrals :\(\int\frac{x\,tan^{-1}x^2}{1+x^4}\)dx
Answer» Assume, tan^-1x²= t d( tan^-1x²) = dt ⇒ \(\frac{2x}{x^4+1}\) = dt ⇒ \(\frac{x}{x^4+1}\) = \(\frac{dt}{2}\) Substituting t and dt ⇒ \(\frac{1}{2}\)\(\int\)t dt ⇒ \(\frac{t^2}{4}\) + c But, t = tan^-1x² ⇒ \(\frac{(tan^{-1}x^2)^2}{4}\) + c.

Evaluate the following integrals :\(\int\frac{x\,tan^{-1}x^2}{1+x^4}\)dx

Answer»

Assume,

tan^-1x²= t

d( tan^-1x²) = dt

⇒ \(\frac{2x}{x^4+1}\) = dt

⇒ \(\frac{x}{x^4+1}\) = \(\frac{dt}{2}\)

Substituting t and dt

⇒ \(\frac{1}{2}\)\(\int\)t dt

⇒ \(\frac{t^2}{4}\) + c

But,

t = tan^-1x²

⇒ \(\frac{(tan^{-1}x^2)^2}{4}\) + c.