Question

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

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

Answer 1

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

Consider, $I=\int \frac{e^{co{t}^{-1}x}}{1+x^2}(x^2-x+1)dx$

$\Rightarrow$ $I=\int e^{co{t}^{-1}x}-\int \frac{x{e}^{co{t}^{-1}x}}{1+x^2}dx$

$\Rightarrow$ $\int e^{co{t}^{-1}x}=x{e}^{co{t}^{-1}x}+\int \frac{x{e}^{co{t}^{-1}x}}{1+x^2}+C$

$\Rightarrow$ $\int e^{co{t}^{-1}x}-\int \frac{x{e}^{co{t}^{-1}x}}{1+x^2}dx=x{e}^{co{t}^{-1}x}+C$

$\Rightarrow I=x{e}^{co{t}^{-1}x}+C$

Hence, the answer is option (B).