Question

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

Answer 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.

Complete step-by-step answer:
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.