Question

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

• A. $x^2{\tan }^{-1}x$
• B. $x+e^{{\tan }^{-1}x}$
• C. $\frac{({\tan }^{-1}x{)}^2}{2}$
• D. $x{e}^{ta{n}^{-1}x}$

Answer 1

The correct option is D $x{e}^{ta{n}^{-1}x}$

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

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

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

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

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

(Using $\int e^t\left(f\right(t)+f^{\prime }(t\left)\right)dt=e^{t}f\left(t\right)+c=e^{\prime }\tan t+c)$

where $f\left(t\right)=\tan t$

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

Answer (d) is correct.