Question Number 217431 by peter frank last updated on 13/Mar/25
Answered by Frix last updated on 13/Mar/25
$$\mathrm{Simply}\:\mathrm{by}\:\mathrm{parts}: \\ $$$${u}'=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\rightarrow\:{u}=−\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$${v}={x}\mathrm{e}^{{x}} \:\rightarrow\:{v}'=\left({x}+\mathrm{1}\right)\mathrm{e}^{{x}} \\ $$$$\int{u}'{v}={uv}−\int{uv}' \\ $$$$\int\frac{{x}\mathrm{e}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }=−\frac{{x}\mathrm{e}^{{x}} }{{x}+\mathrm{1}}+\int\mathrm{e}^{{x}} {dx}=…=\frac{\mathrm{e}^{{x}} }{{x}+\mathrm{1}}+{C} \\ $$
Answered by Spillover last updated on 13/Mar/25
