Question

${\int }_1^2(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx$.

Answer 1

As we can see that

$d(\frac{e^{2x}}{2x})=e^{2x}(\frac{1}{x}-\frac{1}{2x^2})dx$

${\int }_1^{2}d(\frac{e^{2x}}{2x})$

$[\frac{e^{2x}}{2x}{]}_1^2$

$\frac{e^4}{4}-\frac{e^2}{2}$