Question

Integrate the function: $x(logx)^2$

Answer 1

The correct answer is: $\frac{x^2}{2}(logx)^2-\frac{x^2}{2}logx+\frac{x^2}{4}+C$
Let $I=∫x(logx)^2 dx$
Taking$(logx)^2$ as first function and 1 as second function and integrating by parts,we
obtain
$I=(logx)^2∫x.dx-∫[{\frac{d}{dx}logx)^2}∫x dx]dx$
$=\frac{x^2}{2}(logx)^2-[∫2logx.\frac{1}{x}.\frac{x^2}{2}.dx]$
$=\frac{x^2}{2}(logx)^2-∫xlogx. dx$
Again integrating by parts,we obtain
$I=\frac{x^2}{2}(logx)^2-[logx∫x dx-∫[{(\frac{d}{dx}logx)∫x dx}]dx]$
$=\frac{x^2}{2}(logx)^2-[\frac{x^2}{2}-logx-∫\frac{1}{x}.\frac{x^2}{2}dx]$
$=\frac{x^2}{2}(logx)^2-\frac{x^2}{2}logx+\frac{1}{2}∫x dx$
$=\frac{x^2}{2}(logx)^2-\frac{x^2}{2}logx+\frac{x^2}{4}+C$