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Integrate the function\[x^2logx\]

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  • (i)When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts\[\int udv=uv-\int vdu\]
  • (ii)$\int\frac{1}{x}dx=log x+c.$
Given $I=\int x^2log xdx.$
Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts where \[\int udv=uv-\int vdu\]
Let u=log x.
On differentiating with respect to x,
let $dv=x^2 dx.$
On integrating on both sides we get
Now substituting for u,v,du and dv we get,
$\int x^2log xdx=(log x.\frac{x^3}{3})-\int\frac{x^3}{3}.\frac{1}{x}dx.$
On cancelling x we get
$I=(log x.\frac{x^3}{3})-\frac{1}{3}\int x^2dx.$
On integrating we get,
$\;\;\;=\frac{x^3}{3}log x-\frac{1}{3}(\frac{x^3}{3})+c.$
$\;\;\;=\frac{x^3}{3}log x-\frac{x^3}{9}+c.$


answered Feb 8, 2013 by sreemathi.v
edited Feb 8, 2013 by sreemathi.v