Question

Evaluate the integrals using substitution.
${\int }_1^2(\frac{1}{x}-\frac{1}{2x^2})e^{x}dx.$

Answer 1

Let $I={\int }_1^2(\frac{1}{x}-\frac{1}{2x^2})e^{x}dx={\int }_1^2\frac{1}{x}{e}^{2x}dx-{\int }_1^2\frac{1}{2x^2}{e}^{2x}dx$
Evaluate the first integral by rule of integration by parts taking $\frac{1}{x}$ as the first function, we get
$I=[\frac{1}{x}.\frac{e^{2x}}{2}]+{\int }_1^2\frac{1}{x^2}.\frac{e^{2x}}{2}dx-{\int }_1^2\frac{1}{2x^2}dx={\left[\frac{1}{x}\frac{e^{2x}}{2}\right]}_1^2=\frac{e^4}{4}-\frac{e^2}{2}=\frac{e^2}{2}(\frac{e^2}{2}-1)$