Question

If $f\left(x\right)=\int \left(\frac{x^2+{\sin }^{2}x}{1+x^2}\right){\sec }^{2}xdx$ and $f\left(0\right)=0$, then $f\left(1\right)$ equals

• A. $\tan 1-\frac{\pi }{4}$
• B. $\tan 1+1$
• C. $\frac{\pi }{4}$
• D. $1-\frac{\pi }{4}$

Answer 1

The correct option is A $\tan 1-\frac{\pi }{4}$

We have

$f\left(x\right)=\int \left(\frac{x^2+{\sin }^{2}x}{1+x^2}\right){\sec }^{2}xdx$

$=\int \frac{x^2{\sec }^{2}x+{\tan }^{2}x}{1+x^2}dx$

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

$=\tan x-{\tan }^{-1}x+c$

Since,

$f\left(0\right)=0\Rightarrow c=0$

$f\left(1\right)=\tan \left(1\right)-\frac{\pi }{4}$

Hence, the option $\left(A\right)$ is correct answer.