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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\large \frac {\log^{99} x}{x}$

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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 $\large\frac{dx}{dx}=g'(t).$
  • dx=g'(t)dt.
  • Thus $ I=\int f(g(t).g'(t))dt.$
Given $I=\int \frac{(log x)^{99}}{x}dx.$-------(1)
Let us substitute log x=t.
Differentiating on both sides we get,
$\frac{1}{x}dx=dt$.
Now substituting for log x and $\frac{1}{x}dx$ we get,
$I=\int t^{99}.dt$
On integrating we get,
$\frac{t^{100}}{100}+c$.
Substituting back for t we get,
$\frac{1}{100}(log|x|)^{100}+c$
Hence $\int\frac{(log x)^2}{x}dx=\frac{1}{100}(log|x|)^{100}+c$.
answered Dec 24, 2013 by balaji.thirumalai
edited Jan 31, 2014 by yamini.v
 
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