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Integrate the function $\frac{\large x}{\large 9-4x^2}$

1 Answer

  • 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