Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals3 Marks
Question
Evaluate the following integrals:$\int\frac{\log\text{x}}{\text{x}^{\text{n}}}\text{dx}$
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Answer
$\int\frac{\log\text{x}}{\text{x}^{\text{n}}}\text{dx}=\int(\log\text{x})\Big(\frac{1}{\text{x}^{n}}\Big)\text{dx}$
by integration by parts
$\int(\log\text{x})\Big(\frac{1}{\text{x}^{\text{n}}}\Big)\text{dx}=\log\text{x}\int\Big(\frac{1}{\text{x}^{\text{n}}}\Big)\text{dx}-\int\Big(\frac{\text{d}(\log\text{x})}{\text{dx}}\Big)\Big(\int\Big(\frac{1}{\text{x}^{n}}\Big)\text{dx}\Big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)-\int\frac{1}{\text{x}}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)\text{dx}=\log\text{x}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)-\int\Big(\frac{\text{x}^{-\text{n}}}{1-\text{n}}\Big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)-\Big(\frac{1}{1-\text{n}}\Big)\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)=\log\text{x}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)-\bigg(\frac{\text{x}^{1-\text{n}}}{[1-\text{n}]^2}\bigg)+\text{C}$
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