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# Integrate the function$x(log x)^2$

$\begin{array}{1 1} \large \frac{x^2(\log x)^2}{2}-\frac{x^2(\log x)}{2}+\frac{x^2}{4}+c \\ \large \frac{x^2(\log x)^2}{2}+\frac{x^2(\log x)}{2}+\frac{x^2}{4}+c \\\large \frac{x^2(\log x)^2}{2}-\frac{x^2(\log x)}{2}-\frac{x^2}{4}+c \\ \large \frac{x(\log x)^2}{2}-\frac{x(\log x)}{2}+\frac{x^2}{4}+c \end{array}$

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## 1 Answer

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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)$\int \frac{1}{x}dx=log x.$
• (iii)$\frac{d}{dx}(log x)=\frac{1}{x}.$
Given $I=\int x(log x)^2dx.$

Clearly the given integral function is of the form $\int u dv$

Let u=$(log x)^2.$

On differentiating we get,

du=$2(log x).\frac{1}{x}dx.$

Let dv=xdx,On integrating we get,

$v=\frac{x^2}{2}.$

Now substituting for u,v,du and dv we get,

$(log x)^2.(\frac{x^2}{2})-\int \frac{x^2}{2}.2log x.\frac{1}{x}dx.$

Cancelling the common terms,

$\;\;\;=\frac{x^2(log x)^2}{2}-\int xlog xdx.$

Again $\int xlog xdx$ is of the form $\int udv.$

Let u=log x,on differentiating we get,

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

Let dv=xdx,on integrating we get

$v=\frac{x^2}{2}.$

Now substituting for u,v,du and dv we get,

$I_1=\int xlog xdx=\frac{x^2log x}{2}-\int\frac{x^2}{2}\frac{1}{x}dx.$

Cancelling the common terms we get

$I_1=\frac{x^2log x}{2}-\frac{1}{2}\int xdx.$

On integrating we get,

$I_1=\frac{x^2log x}{2}-\frac{1}{2}(\frac{x^2}{2})+c.$

Now substituting for $I_1$ in I we get,

$\int x(log x)^2=\frac{x^2(log x)^2}{2}-\frac{x^2(log x)}{2}+\frac{x^2}{4}+c.$

answered Feb 11, 2013

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