Question

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

• 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. $e^{{\tan }^{-1}x}.{\left({\csc }^{-1}\left(\sqrt{1+x^2}\right)\right)}^2+C$

Answer 1

Answer: (C)

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

Put ${\text{tan}}^{-1}=z$ then it becomes;

$\int \frac{e^{{\text{tan}}^{-1}x}}{(1+x^2)}(({\text{sec}}^{-1}\sqrt{1+x^2}{)}^2+{\text{cos}}^{-1}(\frac{1-x^2}{1+x^2}))dx=\int e^z(z^2+2z)dz$

$=e^z{z}^2+c=e^{{\text{tan}}^{-1}x}({\text{sec}}^{-1}(\sqrt{(}1+x^2)){)}^2+c$