Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\log x+\frac{1}{x^2}\right)e^{x}dx \\ =\int e^x\left(\log x+\frac{1}{x}-\frac{1}{x}+\frac{1}{x^2}\right)dx \\ =\int e^x\left(\log x+\frac{1}{x}\right)dx+\int e^x\left(-\frac{1}{x}+\frac{1}{x^2}\right)dx \\ \overline{)\begin{array}{c}\mathrm{let}{e}^x\log x=t\\ \mathrm{Diff}\mathrm{both}\mathrm{sides}\\ \left(e^x\log x+e^x\frac{1}{x}\right)dx=dt\end{array}}\begin{array}{c}\mathrm{let}{e}^x\left(-\frac{1}{x}\right)=p\\ \mathrm{Diff}\mathrm{both}\mathrm{sides}\\ \left(e^x\frac{-1}{x}+e^x\frac{1}{x^2}\right)dx=dp\end{array} \\ \therefore I=\int dt+\int dp \\ =t+p+C \\ =e^x\log x+e^x\frac{-1}{x}+C \\ =e^x\left(\log x-\frac{1}{x}\right)+C \end{gathered}$$