$∫e^x(logx+\frac{1}{x})dx=$
Answer & explanation
Correct answer: option 1
$∫e^x(logx+\frac{1}{x})dx$ This is equivalent
$∫e^x(f(x)f'(x))dx=e^x.f(x)+C$
$∴∫e^x(logx+\frac{1}{x})dx=e^x.logx+C$
$∫e^x(logx+\frac{1}{x})dx=$
Correct answer: option 1
$∫e^x(logx+\frac{1}{x})dx$ This is equivalent
$∫e^x(f(x)f'(x))dx=e^x.f(x)+C$
$∴∫e^x(logx+\frac{1}{x})dx=e^x.logx+C$