Mathematics

Evaluate the definite integral :
$$\displaystyle \int_{e}^{e^2} \left\{\dfrac {1}{\log x} -\dfrac {1}{(\log x)^2}\right\} dx$$ 


SOLUTION
$$I=\displaystyle \int_e^{e^2}\dfrac {1}{\log x}.1\displaystyle \int_e^{e^2}\dfrac {1}{(\log x)^2}dx$$

                    $$I$$                       $$II$$

For the first part of the integral, taking $$I$$ as the first function and $$II$$ as the second function
  
we have,

$$\left[\dfrac {x}{\log x}\right]_e^{e^2}-\displaystyle \int_e^{e^2}\dfrac {1}{x(\log x)^2}x\ dx-\displaystyle \int_e^{e^2}\dfrac {1}{(\log x)^2}dx$$


$$\therefore \ I=\dfrac {e^2}{\log e^2}-\dfrac {e}{\log e}$$

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

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

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Subjective Medium Published on 17th 09, 2020
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