Question

The value of $\int \frac{(x+1)e^x}{1+x^2{e}^{2x}}\cdot {\tan }^{-1}\left(x{e}^x\right)dx$ is
(where $C$ is constant of integration)

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

Answer 1

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

Given : $I=\int \frac{(x+1)e^x}{1+x^2{e}^{2x}}\cdot {\tan }^{-1}\left(x{e}^x\right)dx$
Putting ${\tan }^{-1}\left(x{e}^x\right)=y$
$\Rightarrow \frac{1}{1+x^2{e}^{2x}}\cdot (x+1)e^{x}dx=dy$
$\Rightarrow I=\int ydy=\frac{y^2}{2}+C$
$=\frac{\left[{\tan }^{-1}\right(x{e}^x){]}^2}{2}+C$