Question

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

• A. $e-\frac{e^2}{2}$
• B. $\frac{e^2}{2}-e$
• C. $\frac{e^2}{2}+e$
• D. $\frac{e^2}{2}-2$

Answer 1

Answer: (B)

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

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

$={\left[\frac{1}{x}{e}^x\right]}_1^2+{\int }_1^2\frac{e^x}{x^2}dx-{\int }_1^2\frac{e^x}{x^2}dx..........$$\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

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