Question

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

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

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