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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following definite integrals:$\int_{\text{e}}^\limits{\text{e}^2}\Big\{\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big\}\text{dx}$
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Answer

Let $\text{I}=\int_{\text{e}}^\limits{\text{e}^2}\Big\{\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big\}\text{dx}$ Then,$\text{I}=\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{\log\text{x}}\text{ dx}-\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})^2}\text{ dx}$
Integrating by parts$\Rightarrow\text{I}=\Bigg\{\Big[\frac{\text{x}}{\log\text{x}}\Big]^{\text{e}^2}_\text{e}-\int_{\text{e}}^\limits{\text{e}^2}\frac{-1}{\text{x}(\log\text{x})^2}\text{x dx}\Bigg\}-\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}}{\log\text{x}}\Big]^{\text{e}^2}_\text{e}+\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})^2}\text{ dx}-\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}}{\log\text{x}}\Big]^{\text{e}^2}_\text{e}+0$
$\Rightarrow\text{I}=\frac{\text{e}^2}{\log\text{e}^2}-\frac{\text{e}}{\log\text{e}}$
$\Rightarrow\text{I}=\frac{\text{e}^2}{2\log\text{e}}-\frac{\text{e}}{\log\text{e}}$
$\Rightarrow\text{I}=\frac{\text{e}^2}{2}-\text{e}$

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