Question

Solve: $\int e^{co{t}^{-1}x}(1-\frac{x}{1+x^2})dx$

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

Answer 1

The correct option is C $x\cdot e^{co{t}^{-1}x}$

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

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

Taking ${\cot }^{-1}x=t$

$\therefore x=\cot t$

$\therefore dx=-cose{c}^{2}t$ dt

$\therefore I=\int e^t\left(\frac{1+{\cot }^{2}t-\cot t}{1+{\cot }^{2}t}\right)\left(cose{c}^{2}t\right)dt$

$=\int e^t(cose{c}^{2}t-cott)dt$

$=\int e^t(\cot t-cose{c}^{2}t)dt$

$\therefore$ Now taking $f\left(t\right)=\cot$

$\therefore f^{\prime }\left(t\right)=-cose{c}^{2}t$

$\therefore t=\int e^t\left[f\right(t)+f^{\prime }(t\left)\right]dt$

$=e^{t}f\left(t\right)+c$

$=e^{{\cot }^{-1}x}\left|\cot \right({\cot }^{-1}x\left)\right|+c$

$=e^t\left|\cot \right({\cot }^{-1}x\left)\right|+c$

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