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Q4  Integrate the functions  x \log x

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Given function is
f(x)=x.\log x
We will use integration by parts method
\int x.\log xdx= \log x.\int xdx - \int(\frac{d(\log x)}{dx}.\int x dx)dx\\ \\ \int x\log xdx = \log x.\frac{x^2}{2}- \int (\frac{1}{x}.\frac{x^2}{2})dx\\ \int x\log xdx = \log x.\frac{x^2}{2}- \int \frac{x}{2}dx\\ \int x\log xdx = \log x.\frac{x^2}{2}- \frac{x^2}{4}+ C

Therefore, the answer is \frac{x^2}{2}\log x- \frac{x^2}{4}+ C
 

Posted by

Gautam harsolia

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