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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\frac{\large x}{\large 9-4x^2}$

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Toolbox:
  • Method of substitution :
  • Given 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 $\frac{dx}{dt}=g'(t).$
  • dx=g'(t)dt.
  • Thus $I=\int f(g(t).g'(t))dt.$
Given $I=\int\frac{x}{9-4x^2}dx$.
 
Let $9-4x^2=t$.
 
$-8xdx=dt\Rightarrow xdx=\frac{-dt}{8}$.
 
Now substituting for x and dx we get,
 
$I=\int\frac{-dt}{8t}=\frac{-1}{t}\int dt.$
 
On integrating we get,
 
$\;\;\;=\frac{-1}{8}log |t|+c.$
 
Substituting for t we get,
 
$\;\;\;\int\frac{x}{9-4x^2}dx=\frac{-1}{8}log |9-4x^2|+c.$
 
Hence $\int\frac{x}{9-4x^2}dx=\frac{-1}{8}log |9-4x^2|+c.$
 
 
 

 

answered Jan 28, 2013 by sreemathi.v
 
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