Question

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

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

Answer 1

Answer: (D)

${xe}^{{\tan }^{-1}x+C}$

${\tan }^{-1}x=t\Rightarrow x=\tan t$

$dx.\frac{1}{1+x^2}=dt.$

$I=\int e^t(1+\tan t+{\tan }^{2}t)dt$

$=\int e^t(\tan t+{\sec }^{2}t)dt[\because \int e^x(f\left(x\right)+f^{\prime }\left(x\right).dx=e^{x}f\left(x\right)+c]$

$=e^t\tan t+c$

$=e^{{\tan }^{-1}x}\tan \left({\tan }^{-1}x\right)+c$

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