Question

$\int \frac{e^{{\tan }^{-1}x}}{(1+x^2)}[{\left({\sec }^{-1}\sqrt{1+x^2}\right)}^2+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)]dx(x>0)$ equals to
(where $c$ is the constant of integration)

• A. $2e^{{\tan }^{-1}x}\cdot {\tan }^{3}x+c$
• B. $\frac{e^x}{{\left({\tan }^{-1}x\right)}^2}+c$
• C. $e^{{\tan }^{-1}x}{\left({\tan }^{-1}x\right)}^2+c$
• D. $e^x\frac{{\left({\tan }^{-1}x\right)}^3}{3}+c$

Answer 1

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

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

We know for $x>0$
${\sec }^{-1}\sqrt{1+x^2}={\tan }^{-1}x{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2{\tan }^{-1}x\Rightarrow I=\int \frac{e^{{\tan }^{-1}x}}{1+x^2}[{\left({\tan }^{-1}x\right)}^2+2{\tan }^{-1}x]dx$
Putting ${\tan }^{-1}x=t$
$\Rightarrow \frac{dx}{1+x^2}=dt=\int e^t(t^2+2t)dt\underset{f\left(x\right)}{\downarrow }\underset{f^{\prime }\left(x\right)}{\downarrow }=(e^t\cdot t^2)+c[\because \int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+c]=e^{{\tan }^{-1}x}{\left({\tan }^{-1}x\right)}^2+c$