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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[x\;log2x\]

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

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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 $I=\int xlog 2xdx.$
 
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 2x.
 
On differentiating with respect to x,
 
$du=2\frac{1}{2x}dx=\frac{1}{x}dx.$
 
let dv=x dx.
 
On integrating we get,
 
$v={x^2}{2}$.
 
On substituting for u,v,du and dv we get,
 
$\int xlog 2xdx=(log 2x.\frac{1}{x^2})-\int\frac{x^2}{2}.\frac{1}{x}dx.$
 
On cancelling x we get
 
$I=\frac{x^2log 2x}{2}-\frac{1}{2}\int x dx.$
 
On integrating we get,
 
$\int xlog 2xdx=\frac{x^2log 2x}{2}-\frac{x^2}{4}+c.$

 

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