Question

$\int \frac{2x{\tan }^{-1}{x}^2}{1+x^4}dx$ is equal to :

• A. $[{\tan }^{-1}{x}^2{]}^2+C$
• B. $\frac{1}{2}[{\tan }^{-1}{x}^2{]}^2+C$
• C. $2[{\tan }^{-1}{x}^2{]}^2+C$
• D. None of the above.

Answer 1

Answer: (B)

$\frac{1}{2}[{\tan }^{-1}{x}^2{]}^2+C$

Let $I=\int \frac{2x{\tan }^{-1}{x}^2}{1+x^4}dx$

Put $t={\tan }^{-1}{x}^2$
$\Rightarrow dt=\frac{2x}{1+x^4}dx$

$\therefore \int \frac{2x{\tan }^{-1}{x}^2}{1+x^4}dx=\int tdt$

$=\frac{t^2}{2}+C$

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