Question

$\int \frac{e^{{\cot }^{-1}x}}{1+x^2}(x^2-x+1)dx=$ _____ $+c$

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

Answer 1

The correct option is B $x\cdot e^{{\cot }^{-1}x}$

$\int e^{{\cot }^{-1}x}\left(\frac{x^2+1-x}{1+x^2}\right)dx=\int e^{{\cot }^{-1}x}(1-\frac{x}{1+x^2})dx$
$=\int e^{{\cot }^{-1}x}dx-\int e^{{\cot }^{-1}x}\left(\frac{x}{1+x^2}\right)dx$
$=e^{{\cot }^{-1}x}\cdot x-\int -e^{{\cot }^{-1}x}\left(\frac{1}{1+x^2}\right)\cdot xdx-\int e^{{\cot }^{-1}x}\left(\frac{x}{1+x^2}\right)dx$
$=e^{{\cot }^{-1}x}\cdot x+C$