Question

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

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

Answer 1

Answer: (A)

Let $I=\int e^x\left(\frac{x^2{\tan }^{-1}x+{\tan }^{-1}x+1}{x^2+1}\right)dx$
$\Rightarrow I=\int e^x\left({\tan }^{-1}x+\frac{1}{x^2+1}\right)dx$
If $f(x)={\tan }^{-1}x$, then
$f^{\prime }(x)=\frac{1}{x^2+1}$
$\therefore I=\int e^x\left({\tan }^{-1}x+\frac{1}{x^2+1}\right)dx$
$=e^x{\tan }^{-1}x+C$
$\left[\because \int e^x\left[f(x)+f^{\prime }(x)\right]dx=e^{x}f(x)+C\right]$