# Integrate the function$x^2logx$

## 1 Answer

Toolbox:
• (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,

$du=\frac{1}{x}dx.$

let $dv=x^2 dx.$

On integrating on both sides we get

$v=\frac{x^3}{3}.$

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
edited Feb 8, 2013

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