Question

If $\int \frac{x^2-x+1}{x^2+1}{e}^{{\cot }^{-1}x}dx=f\left(x\right)\cdot e^{{\cot }^{-1}x}+C$, then $f\left(x\right)$ is equal to:
(where $C$ is integration constant)

Answer 1

Let $I=\int \frac{x^2-x+1}{x^2+1}\cdot e^{{\cot }^{-1}x}dx$
Put $x=\cot t$
$\Rightarrow -{\text{cosec}}^{2}tdt=dx$
Now,
$\Rightarrow I=\int \frac{e^t({\cot }^{2}t-\cot t+1)}{(1+{\cot }^{2}t)}(-{\text{cosec}}^{2}t)dt=-\int e^t({\text{cosec}}^{2}t-\cot t)dt=\int e^t(\cot t-{\text{cosec}}^{2}t)dt\underset{f\left(x\right)}{\downarrow }\underset{f^{\prime }\left(x\right)}{\downarrow }=e^t\cot t+C[\because \int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+C]=x\cdot e^{{\cot }^{-1}x}+C\equiv f\left(x\right)\cdot e^{{\cot }^{-1}x}+C\therefore f\left(x\right)=x$