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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