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

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

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 $I=\int\frac{1}{x(log x)^m}dx,x>0$.

Let $log x=t$.

$\;\;\;\frac{1}{x}dx=dt$.

Now substituting for x and dx we get,

$I=\int\frac{dt}{t^m}$.

On integrating we get,
$\;\;\;=\frac{t^{-m+1}}{-m+1}+c.$

Substituting for m we get,

$\;\;\;=\frac{(log x)^{1-m}}{1-m}+c.$

Hence $\int\frac{1}{x(log x)^m}dx=\frac{(log x)^{1-m}}{1-m}+c.$