Question

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

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

Answer 1

The correct option is E

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

Explanation for the correct option:

Find the value of given integral.

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

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

$$\begin{gathered} =\int e^{{\tan }^{-1}x}\left[\frac{x}{1+x^2}+1\right]dx \\ =\int \left(e^{{\tan }^{-1}x}\frac{x}{1+x^2}+e^{{\tan }^{-1}x}\right)dx \\ =\int \left(\frac{d}{dx}x{e}^{{\tan }^{-1}x}\right)dx \end{gathered}$$

$=x{e}^{{\tan }^{-1}x}+c$, Where $c$is the integration constant.

Hence, option (E) is the correct answer.