Question

$\int \frac{e^{{\tan }^{-1}x}}{(1+x^2)}[{\left({\sec }^{-1}\sqrt{1+x^2}\right)}^2+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx]$ at $(x>0)$

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

Answer 1

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

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

Substitute

${\tan }^{-1}\left(x\right)=t$ and $\frac{dx}{1+x^2}=dt$

We get,

$I=\int e^t[{\left(t\right)}^2+2t]dt=\int t^2{e}^{t}dt+\int 2t{e}^{t}dt$

By integration by parts we get

$I=t^2{e}^t-\int 2t{e}^{t}dt+\int 2t{e}^{t}dt+C{=t}^2{e}^t+C=e^{{\tan }^{-1}\left(x\right)}{\left({\sec }^{-1}\sqrt{1+x^2}\right)}^2+C$

Hence, option 'C' is correct.