Mathematics

# $\displaystyle \int_{1}^{2}e^{x}\left ( \dfrac{1}{x}-\dfrac{1}{x^{2}} \right )\ dx$ equals to

##### ANSWER

$e\ \left ( \dfrac{e}{2}-1 \right )$

##### SOLUTION
Now,
$\displaystyle\int\limits_{1}^{2}e^x\left(\dfrac{1}{x}-\dfrac{1}{x^2}\right)dx$

$=\displaystyle\int\limits_{1}^{2}e^x\left(\dfrac{1}{x}\right)dx$$-\displaystyle\int\limits_{1}^{2}e^x\left(\dfrac{1}{x^2}\right)dx =\left.\left(\dfrac{e^x}{x}\right)\right|_{1}^{2}$$+\displaystyle\int\limits_{1}^{2}e^x\left(\dfrac{1}{x^2}\right)dx$$-\displaystyle\int\limits_{1}^{2}e^x\left(\dfrac{1}{x^2}\right)dx$ [ Using method of by parts]

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

$=e\left(\dfrac{e}{2}-1\right)$.

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Single Correct Medium Published on 17th 09, 2020
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