$∫e^x(logx+\frac{1}{x})dx=$

ex log x + C $\frac{e^x}{x}+C$ x log x + C $\frac{e^xlogx}{x}+C$

ex log x + C

$∫e^x(logx+\frac{1}{x})dx$ This is equivalent  $∫e^x(f(x)f'(x))dx=e^x.f(x)+C$ $∴∫e^x(logx+\frac{1}{x})dx=e^x.logx+C$