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
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