Question

Integrate $\int e^x(\frac{1}{x}-\frac{1}{x^2}).dx$

Answer 1

Property of integration

$\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+c$

Let $f\left(x\right)=\frac{1}{x}$

Differentiating $f^{\prime }\left(x\right)=-\frac{1}{x^2}$

$\therefore \int e^x(\frac{1}{x}-\frac{1}{x^2})dx=\int e^x[\frac{1}{x}+(-\frac{1}{x^2}\left)\right]dx$

$=e^x.\frac{1}{x}+c$

$\int e^x(\frac{1}{x}-\frac{1}{x^2})dx=\frac{e^x}{x}+c$