Question

$\underset{1}{\overset{2}{\int }}{e}^{2x}\left(\frac{1}{x}-\frac{1}{2x^2}\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I{\int }_1^2{e}^{2x}\left(\frac{1}{x}-\frac{1}{2x^2}\right)\mathit{d}x.\mathrm{Then}, \\ I={\int }_1^2{e}^{2x}\frac{1}{x}-{\int }_1^2{e}^{2x}\frac{1}{2x^2}dx \\ \mathrm{Integrating}\mathrm{first}\mathrm{term}\mathrm{by}\mathrm{parts} \\ \Rightarrow I=\left\{{\left[\frac{e^{2x}}{2x}\right]}_1^2-{\int }_1^2-e^{2x}\frac{1}{2x^2}\right\}-{\int }_1^2{e}^{2x}\frac{1}{2x^2}dx \\ \Rightarrow I={\left[\frac{e^{2x}}{2x}\right]}_1^2 \\ \Rightarrow I=\frac{e^4}{4}-\frac{e^2}{2} \\ \Rightarrow I=\frac{e^4-2e^2}{4} \end{gathered}$$