Question
Evaluate the following integrals:$\int\log_{10}\text{x dx}$
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Answer
Let $\text{I}=\int\log_{10}\text{x dx}$
$=\int\frac{\log\text{x}}{\log10}\text{dx}$
$=\frac{1}{\log10}\int1\times\log\text{x dx}$
Using integration by parts,
$=\frac{1}{\log10}\Big[\log\text{x}\int\text{dx}-\int\Big(\frac{1}{\text{x}}\int\text{dx}\Big)\text{dx}\Big]$
$=\frac{1}{\log10}\Big[\text{x}\log\text{x}-\int\big(\frac{\text{x}}{\text{x}}\big)\text{dx}\Big]$
$=\frac{1}{\log10}[\text{x}\log\text{x}-\text{x}]$
$\text{I}=\frac{\text{x}}{\log10}(\log\text{x}-1)$
$=\int\frac{\log\text{x}}{\log10}\text{dx}$
$=\frac{1}{\log10}\int1\times\log\text{x dx}$
Using integration by parts,
$=\frac{1}{\log10}\Big[\log\text{x}\int\text{dx}-\int\Big(\frac{1}{\text{x}}\int\text{dx}\Big)\text{dx}\Big]$
$=\frac{1}{\log10}\Big[\text{x}\log\text{x}-\int\big(\frac{\text{x}}{\text{x}}\big)\text{dx}\Big]$
$=\frac{1}{\log10}[\text{x}\log\text{x}-\text{x}]$
$\text{I}=\frac{\text{x}}{\log10}(\log\text{x}-1)$
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