Question

# Integrate the function $\dfrac{1}{x{{\left( \log x \right)}^{m}}},x>0,m\ne 1$

Hint:Substitute $t=\log x$, differentiate them and put the values in the integral. Simplify the integral given by the basic integral formula. Replace $t$ with $\log x$ in the simplified integral.

Let’s consider $I=\int{\dfrac{1}{x{{\left( \log x \right)}^{m}}}dx\ldots \ldots (1)}$
Let’s assume that $\log x=t.$
\begin{align} & \therefore x={{\log }^{-1}}t={{e}^{t}} \\ & \Rightarrow x={{e}^{t}} \\ \end{align}
Differentiate $\log x=t$ with respect to $x.$
\begin{align} & \log x=t \\ & \Rightarrow \dfrac{1}{x}dx=dt \\ \end{align}
Now substitute these values in equation (1).
\begin{align} & I=\int{\dfrac{1}{x{{\left( \log x \right)}^{m}}}dx=\int{\dfrac{dt}{{{t}^{m}}}}} \\ & I=\int{\dfrac{dt}{{{t}^{m}}}}=\int{{{t}^{-m}}dt\ldots \ldots (2)} \\ \end{align}
Now integrate equation (2).
We know $\int{{{x}^{1}}dx=\dfrac{{{x}^{1+1}}}{1+1}=\dfrac{{{x}^{2}}}{2}}+c$ where c is a constant of integration.
Similarly, $I=\int{{{t}^{-m}}=\left( \dfrac{{{t}^{-m+1}}}{-m+1} \right)+c}$
$I=\dfrac{{{t}^{1-m}}}{1-m}+c$
Now replace $t$ with $\log x$, we get
\begin{align} & I=\dfrac{{{\left( \log x \right)}^{1-m}}}{1-m}+c \\ & I=\dfrac{{{\left( \log x \right)}^{1}}\times {{\left( \log x \right)}^{-m}}}{\left( 1-m \right)}+c \\ & I=\dfrac{1}{1-m}\times \dfrac{\log x}{{{\left( \log x \right)}^{m}}}+c \\ \end{align}

Note: The integration can be solved by basic integral formula $\int{{{x}^{1}}dx=\dfrac{{{x}^{1+1}}}{1+1}=\dfrac{{{x}^{2}}}{2}}+c$
Similarly $\Rightarrow \int{1.dx=x+c;\int{a.dx=ax+c}}$ etc.