0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# The value of the integral $\underset{1}{\overset{2}{\int }}{e}^{x}\left(\frac{1}{x}-\frac{1}{{x}^{2}}\right)dx$ is _______________.

Open in App
Solution

## $\mathrm{Let}I=\underset{1}{\overset{2}{\int }}{e}^{x}\left(\frac{1}{x}-\frac{1}{{x}^{2}}\right)dx\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know},\phantom{\rule{0ex}{0ex}}\int {e}^{x}\left(f\left(x\right)+f\text{'}\left(x\right)\right)dx={e}^{x}f\left(x\right)+c\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Here},\phantom{\rule{0ex}{0ex}}f\left(x\right)=\frac{1}{x}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=-\frac{1}{{x}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\phantom{\rule{0ex}{0ex}}I={\left[{e}^{x}\left(\frac{1}{x}\right)\right]}_{1}^{2}\phantom{\rule{0ex}{0ex}}=\left[{e}^{2}\left(\frac{1}{2}\right)-{e}^{1}\left(\frac{1}{1}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{{e}^{2}}{2}-e\phantom{\rule{0ex}{0ex}}=\frac{{e}^{2}-2e}{2}\phantom{\rule{0ex}{0ex}}=\frac{e\left(e-2\right)}{2}$ ​ Hence, the value of the integral $\underset{1}{\overset{2}{\int }}{e}^{x}\left(\frac{1}{x}-\frac{1}{{x}^{2}}\right)dx$ is $\overline{)\frac{e\left(e-2\right)}{2}}.$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Properties of Inequalities
MATHEMATICS
Watch in App
Join BYJU'S Learning Program