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Q35  Integrate the functions  \frac{( 1+ \log x )^2}{x}

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Given function  \frac{( 1+ \log x )^2}{x}

Assume that 1+\log x =t

\therefore \frac{1}{x}dx =dt

= \int \frac{(1+\log x )^2}{x}dx = \int t^2 dt

= \frac{t^3}{3}+C

Now, back substituting the value of t,

= \frac{(1+\log x )^3}{3}+C

 

 

Posted by

Divya Prakash Singh

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