Question

$\int e^{ta{n}^{-1}x}\left[\frac{1+x+x^2}{1+x^2}\right]dx=$

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

Answer 1

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

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

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

$=\int [e^{{\tan }^{-1}x}+\frac{x{e}^{{\tan }^{-1}x}}{1+x^2}]dx$
For $f\left(x\right)=e^{{\tan }^{-1}x}$, above integral $I$ reduces in the form of $\int \left[f\right(x)+x{f}^{{}^{\prime }}(x\left)\right]dx$

But we know that, $\frac{d}{dx}\left(xf\right(x)+c)=\left[f\right(x)+x{f}^{{}^{\prime }}(x\left)\right]$
$\therefore \int \left[f\right(x)+x{f}^{{}^{\prime }}(x\left)\right]dx=xf\left(x\right)+c$

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

$\therefore I=\int e^{{\tan }^{-1}x}\left[\frac{1+x+x^2}{1+x^2}\right]dx=x{e}^{{\tan }^{-1}x}+c$
Hence, option 'B' is correct.