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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\frac{\large 1}{\large x+x\;log x}$

$\begin{array}{1 1} \log\;|(1+\log x)|+c. \\ \log\;|(1-\log x)|+c. \\ \log\;|(x+\log x)|+c. \\ \log\;|(x-\log x)|+c.\end{array} $

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1 Answer

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Toolbox:
  • Method of substitution:
  • Given $\int f(x)dx$ can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dx}=g'(t).
  • dx=g'(t)dt.
  • Thus $ I=\int f(g(t).g'(t))dt.$
Given $I=\int \frac{1}{x+xlog x}dx=\int\frac{1}{x(1+log x)}dx.$
 
Let (1+log x) be t.
 
Differentiating on both sides we get,
 
$\frac{1}{x}dx=dt$.
 
Now substituting for (1+log x) and $\frac{1}{x}dx$ we get,
 
$I=\int \frac{1}{t}.dt$
 
On integrating we get,
 
log|t|+c.
 
Substituting back for t we get,
 
log|(1+log x)|+c.
 
Hence $\int \frac{1}{x+xlog x}dx=log|(1+log x)|+c.$

 

answered Jan 28, 2013 by sreemathi.v
 
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