Question

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

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

Answer 1

Answer: (A)

$\int \frac{e^{{\cot }^{-1}x}}{1+x^2}\left(x^2-x+1\right)dx$
Let ${\cot }^{-1}x=t$
$\Rightarrow -\frac{1}{1+x^2}dx=dt$
$\int e^t\left(\cot t-{\text{cosec}}^{2}t\right)dt$
$e^t\cot t+c$
$x{e}^{{\cot }^{-1}x}+c$