Mathematics

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

##### SOLUTION
$\displaystyle I = \int_{e}^{{e}^{2}}{\left( \cfrac{1}{\log{x}} - \cfrac{1}{{\left( \log{x} \right)}^{2}} \right) dx}$
Let

$\log{x} = t$
$\Rightarrow x = {e}^{t}$

$\Rightarrow dx = {e}^{t} dt$

Therefore,
$\displaystyle I = \int_{1}^{2}{\left( \cfrac{1}{t} - \cfrac{1}{{t}^{2}} \right) {e}^{t} dt}$
As we know that,

$\displaystyle \int{{e}^{x} \left( f{\left( x \right)} - f^{\prime}{\left( x \right)} \right) dx} = {e}^{x} f{\left( x \right)}$

Therefore,
$\displaystyle I = \int_{1}^{2}{\left( \cfrac{1}{t} - \cfrac{1}{{t}^{2}} \right) {e}^{t} dt} = \left[ \cfrac{{e}^{t}}{t} \right]_{1}^{2} = \left( \cfrac{{e}^{2}}{2} - e \right)$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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