$\int x \tan^{- 1} (x^{2}) d x =$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$x^{2} \tan^{- 1} (x^{2}) - \frac{1}{2} \log (1 + x^{4}) + c$

$\frac{1}{2} [x^{2} \tan^{- 1} (x^{2}) - \frac{1}{2} \log (1 + x^{4})] + c$

$x^{2} \tan^{- 1} (x^{2}) + \frac{1}{2} \log (1 + x^{4}) + c$

$\frac{1}{2} [x^{2} \tan^{- 1} (x^{2}) - \log (1 + x^{4})] + c$

answer is D.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\int \tan^{- 1} x^{2} . x d x$

$= \tan^{- 1} x^{2} \frac{x^{2}}{2} - \int \frac{1}{1 + {(x^{2})}^{2}} (2 x) \frac{x^{2}}{2} d x$

$= \frac{x^{2}}{2} \tan^{- 1} x^{2} - \int \frac{x^{3}}{1 + x^{4}} d x$

$= \frac{x^{2}}{2} \tan^{- 1} x^{2} - \frac{1}{4} \int \frac{4 x^{3}}{1 + x^{4}} d x$

$= \frac{x^{2}}{2} \tan^{- 1} x^{2} - \frac{1}{4} \log | 1 + x^{4} | + c$

$= \frac{1}{2} [x^{2} \tan^{- 1} x^{2} - \frac{1}{2} \log (1 + x^{4})] + c$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test