# Finding integral of x tan^(-1) x dx

• Feb 7th 2011, 01:48 AM
helloying
Finding integral of x tan^(-1) x dx
Find $\displaystyle \frac{d}{dx} [tan^{-1} (x^2)]$ I found it to be $\displaystyle \frac{1}{1+x^4}$

Hence find $\displaystyle \int x tan^{-1} (x^2) dx$

i do integration by parts with u=$\displaystyle tan^{-1} (x^2)$
$\displaystyle \frac{du}{dx}=\frac{1}{x^4+1}$
v=$\displaystyle \frac{x^2}{2}$

Then i got $\displaystyle 0.5x^2 tan^{-1} (x^2) - \int\frac{x^2}{2x^4 +2}$

i got stuck here. pls help!
• Feb 7th 2011, 02:01 AM
tonio
Quote:

Originally Posted by helloying
Find $\displaystyle \frac{d}{dx} [tan^{-1} (x^2)]$ I found it to be $\displaystyle \frac{1}{1+x^4}$

This is wrong. Read and apply correctly the chain rule and try again.

Tonio

Hence find $\displaystyle \int x tan^{-1} (x^2) dx$

i do integration by parts with u=$\displaystyle tan^{-1} (x^2)$
$\displaystyle \frac{du}{dx}=\frac{1}{x^4+1}$
v=$\displaystyle \frac{x^2}{2}$

Then i got $\displaystyle 0.5x^2 tan^{-1} (x^2) - \int\frac{x^2}{2x^4 +2}$

i got stuck here. pls help!

.