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Integrate the function $\frac{\large 1}{\large x(log x)^m},x\;>\;0$

$\begin{array}{1 1} \frac{(\log x)^{1-m}}{1-m}+c \\\frac{(\log x)^{1+m}}{1-m}+c \\ \frac{(\log x)^{1-m}}{1+m}+c \\ \frac{(\log x)^{1+m}}{1+m}+c\end{array}$

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 $I=\int\frac{1}{x(log x)^m}dx,x>0$.
Let $log x=t$.
Now substituting for x and dx we get,
On integrating we get,
Substituting for m we get,
$\;\;\;=\frac{(log x)^{1-m}}{1-m}+c.$
Hence $\int\frac{1}{x(log x)^m}dx=\frac{(log x)^{1-m}}{1-m}+c.$


answered Jan 28, 2013 by sreemathi.v