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Home  >>  CBSE XII  >>  Math  >>  Integrals
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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 by sreemathi.v
 
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