Question

Evaluate :

$\int \frac{{\sec }^2\left(2{\tan }^{-1}x\right)}{1+x^2}dx$

• A. $\frac{1}{3}\tan \left(2{\tan }^{-1}x\right)+x+C$
• B. $\frac{1}{2}{x}^2+C$
• C. $\frac{1}{2}\tan \left(2{\tan }^{-1}x\right)+C$
• D. None of these

Answer 1

The correct option is C $\frac{1}{2}\tan \left(2{\tan }^{-1}x\right)+C$

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

Let $2{\tan }^{-1}x=t$. Then,

$d\left(2{\tan }^{-1}x\right)=dt$

$\frac{2}{1+x^2}dx=dt$

$dx=\frac{1+x^2}{2}dt$

Putting $2{\tan }^{-1}x=t$ and $dx=\frac{1+x^2}{2}dt$, we get,

$I=\int \frac{{\sec }^{2}t}{1+x^2}\times \frac{1+x^2}{2}dt$

$I=\frac{1}{2}\int {\sec }^{2}tdt$

$I=\frac{1}{2}\tan t+C$

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