Question

$\int e^x\frac{x^3-x+2}{{\left(x^2+1\right)}^2}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int e^x\frac{x^3-x+2}{{\left(x^2+1\right)}^2}dx \\ =\int e^x\frac{x^3+x^2-x^2-2x+x+2}{{\left(x^2+1\right)}^2}dx \\ =\int e^x\left[\frac{x^3+x^2-x^2-2x+x+2}{{\left(x^2+1\right)}^2}\right]dx \\ =\int e^x\left[\frac{x^3+x^2+x+1-x^2-2x+1}{{\left(x^2+1\right)}^2}\right]dx \\ =\int e^x\left[\frac{x^3+x^2+x+1}{{\left(x^2+1\right)}^2}+\frac{-x^2-2x+1}{{\left(x^2+1\right)}^2}\right]dx \\ =\int e^x\left[\frac{\left(x^2+1\right)\left(x+1\right)}{{\left(x^2+1\right)}^2}+\frac{-x^2-2x+1}{{\left(x^2+1\right)}^2}\right]dx \\ =\int e^x\left[\frac{\left(x+1\right)}{\left(x^2+1\right)}+\frac{-x^2-2x+1}{{\left(x^2+1\right)}^2}\right]dx \\ \mathrm{As}\mathrm{we}\mathrm{know}\int e^x\left[f\left(x\right)+f\text{'}\left(x\right)\right]dx=e^{x}f\left(x\right)+C \\ \mathrm{Here},\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left[\frac{\left(x+1\right)}{\left(x^2+1\right)}\right]=\frac{-x^2-2x+1}{{\left(x^2+1\right)}^2} \\ \mathrm{Therefore},\mathrm{integral}\mathrm{becomes} \\ I=e^x\left(\frac{x+1}{x^2+1}\right)+C \end{gathered}$$