Question

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

Answer 1

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

$=\int e^x(\log x+\frac{1}{x})dx$

$=\int e^x(\log x+(\log x{)}^{\prime })dx=e^x\log x+C$

Fact: $\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+C$