Question

Solve: $\int \frac{2x.{\tan }^{-1}\left(x^2\right)}{1+x^4}dx$

• A. $({\tan }^{-1}{x}^2{)}^2+C$
• B. $\frac{({\tan }^{-1}{x}^2{)}^2}{2}+C$
• C. $\frac{({\tan }^{-1}{x}^2{)}^2}{2}+C$
  
• D. $\left({\tan }^{-1}{x}^2\right)+C$

Answer 1

The correct option is B

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

Let $I=\int \frac{2x\cdot {\tan }^{-1}\left(x^2\right)}{1+x^4}dx$

Put $x^2=t$ $\Rightarrow 2xdx=dt$

Substituting above values, we have

$I=\int \frac{{\tan }^{-1}\left(t\right)}{1+t^2}dt$
Let ${\tan }^{-1}t=u$ $\Rightarrow \frac{1}{1+t^2}dt=du$

$I=\int udu$

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

$=\frac{1}{2}\left({\tan }^{-1}\right(t){)}^2+C$

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