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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[x\;logx\]

$\begin{array}{1 1} \large \frac{\log x}{x^2}-\frac{x^2}{4}+c \\ \frac{\log x}{x^2}+\frac{x^2}{4}+c \\ \frac{\log x}{x}-\frac{x^2}{4}+c \\ \frac{\log x}{x}-\frac{x}{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)$\int\frac{1}{x}dx=log x+c.$
Given $I=\int xlog xdx.$
 
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 we get
 
$du=\frac{1}{x}dx.$
 
let dv=x dx.
 
On integrating we get
 
$v=\frac{x^2}{2}$.
 
On substituting for u,v,du and dv we get,
 
$\int xlog x=(log x.\frac{1}{x^2})-\int\frac{x^2}{2}.\frac{1}{x}dx.$
 
$\;\;\;\;=\frac{log x}{x^2}-\frac{1}{2}\int xdx.$
 
On integrating we get,
 
$\;\;\;\;=\frac{log x}{x^2}-\frac{1}{2}{x^2}{2}+c.$
 
$\;\;\;=\frac{log x}{x^2}-\frac{x^2}{4}+c.$

 

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