Question

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

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

Answer 1

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

Consider the given equation.

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

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

Let $t=x{e}^{{\tan }^{-1}x}$

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

$\frac{dt}{dx}=e^{{\tan }^{-1}x}\times 1+x\left(e^{{\tan }^{-1}x}\right)\times \frac{1}{1+x^2}$

$\frac{dt}{dx}=e^{{\tan }^{-1}x}+\frac{x{e}^{{\tan }^{-1}x}}{1+x^2}$

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

From equation (1), we get

$I=\int 1dt$

$I=t+C$

On putting the value of $t$, we get

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

Hence, this is the answer.