Hi . Can anyone give me a way to solve this Integral :
$\displaystyle
\int \frac{x^2}{1+x^4} dx
$
Regards ...
EDIT:: Made a mistake. Re-checking everything...
Sub x^2 = tan theta.
x = sqrt(tan theta)
$\displaystyle dx = \frac12 (tan \theta)^{-\frac12}.sec^2\theta$
$\displaystyle \int \frac{x^2}{1+x^4} dx = \int \frac{tan\theta}{1+tan^2 \theta} . \frac12 (tan \theta)^{-\frac12}.sec^2\theta d\theta$
$\displaystyle \int \frac{tan\theta}{1+tan^2 \theta} . \frac12 (tan \theta)^{-\frac12}.sec^2\theta d\theta= \frac12 \int \frac{tan\theta}{sec^2\theta} . (tan \theta)^{-\frac12}.sec^2\theta d\theta$
$\displaystyle \frac12 \int \frac{tan\theta}{sec^2\theta} . (tan \theta)^{-\frac12}.sec^2\theta d\theta= \frac12 \int tan\theta(tan \theta)^{-\frac12} d\theta = \frac12 \int (tan\theta)^{\frac12} d\theta$
$\displaystyle \frac12 \int (tan\theta)^{\frac12} d\theta = \frac12 [\frac{2(tan\theta)^{\frac32}}{3sec^2\theta}]$
Not sure if that's possible...