Mathematics - Indefinite Integration Question with Solution | TestHub

Let $\mathrm{I}=\int \frac{x^2-x+1}{x^2+1}\cdot e^{{\cot }^{-1}x}dx$ Put $x=\cot t\Rightarrow -{\mathrm{cosec}}^{2}tdt=dx$ Now, $1+{\cot }^{2}t={\mathrm{cosec}}^{2}t$ $\therefore$ $\mathrm{I}=\int \frac{e^t\left({\cot }^{2}t-\cot t+1\right)}{\left(1+{\cot }^{2}t\right)}\left(-{\mathrm{cosec}}^{2}t\right)dt$ $$\begin{array}{rl}& =-\int e^t\left(\mathrm{cosec}{}^{2}t-\cot t\right)dt\\ & =\int e^t\left(\cot t-{\mathrm{cosec}}^{2}t\right)dt\\ & =e^t\cot t+\mathrm{C}\\ & =e^{{\cot }^{-1}x}(x)+\mathrm{C}\equiv \mathrm{A}(x)\cdot e^{{\cot }^{-1}x}+\mathrm{C}\\ \Rightarrow \mathrm{A}(x)& =x\end{array}$$