Question

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

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

Answer 1

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

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

$=\int e^x({\tan }^{-1}x+\frac{1}{1+x^2})dx=e^x{\tan }^{-1}x+c$

Note: $\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+c$

Here $f\left(x\right)={\tan }^{-1}x$