Question
Evaluate the following integrals:$\int(\text{x}+1)\log\text{x dx}$
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Answer
$\int(\text{x}+1).\log\text{x dx}$
$=\log \text{x}\int(\text{x}+1)\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\log\text{x})\int(\text{x}+1)\text{dx}\Big\}\text{dx}$
$=\log\text{x}\Big[\frac{\text{x}^2}{2}+\text{x}\Big]-\int\frac{1}{\text{x}}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)-\int\big(\frac{\text{x}}{2}+1\big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)-\Big(\frac{\text{x}^2}{4}+\text{x}\Big)+\text{C}$
$=\log \text{x}\int(\text{x}+1)\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\log\text{x})\int(\text{x}+1)\text{dx}\Big\}\text{dx}$
$=\log\text{x}\Big[\frac{\text{x}^2}{2}+\text{x}\Big]-\int\frac{1}{\text{x}}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)-\int\big(\frac{\text{x}}{2}+1\big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)-\Big(\frac{\text{x}^2}{4}+\text{x}\Big)+\text{C}$
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