Question

The value of integral $\int \frac{e^{{\tan }^{-1}x}}{1+x^2}[{\left({\sec }^{-1}\sqrt{1+x^2}\right)}^2+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)]dx$ $(x>0),$ is
(where ${}^{\prime }{C}^{\prime }$ is the constant of integration)

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

Answer 1

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

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

Put $x=\tan \theta$
${\sec }^{-1}\sqrt{1+x^2}={\sec }^{-1}\sqrt{1+{\tan }^2\theta }={\sec }^{-1}\left(\sec \theta \right)(\because x>0)=\theta$

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

$\Rightarrow I=\int e^{\theta }({\theta }^2+2\theta )d\theta$
The above integral is in the form of
$\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+C$

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