Question

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

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

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

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

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

$=e^x{\tan }^{-1}x+c$