Question

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

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

Answer 1

Answer: (C)

note that ${\sec }^{-1}\sqrt{1+x^2}={\tan }^{-1}x$
${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2{\tan }^{-1}x$ for $x>0$
$I=\int \frac{e^{{\tan }^{-1}x}}{1+x^2}\left({\left({\tan }^{-1}x\right)}^2+2{\tan }^{-1}x\right)dx$ put ${\tan }^{-1}x=t$
$=\int e^t\left(t^2+2t\right)dt=e^t\cdot t^2=e^{{\tan }^{-1}x}{\left({\tan }^{-1}x\right)}^2+C$