Question

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

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

Answer 1

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

$\int e^x(ta{n}^{-1}x+\frac{1}{1+x^2})dx=e^{x}ta{n}^{-1}x+c$
Since, $\int e^x(f\left(x\right)+f^{{}^{\prime }}\left(x\right))dx=e^{x}f\left(x\right)+c$
Ans: C