Question

The value of the integral $\int \frac{e^{2{\tan }^{-1}x}(1+x{)}^2}{1+x^2}dx$ is equal to

• A. $x{e}^{{\tan }^{-1}x}+c$
• B. $x{e}^{2{\tan }^{-1}x}+c$
• C. $2x{e}^{{\tan }^{-1}x}+c$
• D. $2x{e}^{2{\tan }^{-1}x}+c$

Answer 1

The correct option is B $x{e}^{2{\tan }^{-1}x}+c$

$\int \frac{e^{2{\tan }^{-1}x}(1+x{)}^2}{1+x^2}dx$
Putting $2{\tan }^{-1}x=t$
$\Rightarrow \frac{2}{1+x^2}dx=dtI=\frac{1}{2}\int {(1+\tan \frac{t}{2})}^2{e}^{t}dt=\frac{1}{2}\int e^t(2\tan \frac{t}{2}+{\sec }^2\frac{t}{2})dt\underset{f\left(t\right)}{\downarrow }\underset{f^{\prime }\left(t\right)}{\downarrow }=\frac{1}{2}{e}^{t}f\left(t\right)+c=\frac{1}{2}{e}^t\cdot 2\tan \left(\frac{t}{2}\right)+c=e^{2{\tan }^{-1}x}\cdot \tan \left(\frac{2{\tan }^{-1}x}{2}\right)+c=x{e}^{2{\tan }^{-1}x}+c[\because \tan ({\tan }^{-1}x)=x]$