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 $\log x=t.$
$\begin{aligned} & \therefore x={{\log }^{-1}}t={{e}^{t}} \\\ & \Rightarrow x={{e}^{t}} \\\ \end{aligned}$
Differentiate $\log x=t$ with respect to $x.$
$\begin{aligned} & \log x=t \\\ & \Rightarrow \dfrac{1}{x}dx=dt \\\ \end{aligned}$
Now substitute these values in equation (1).

& 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{aligned}$$ Now integrate equation (2). We know $$\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{aligned} & 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{aligned}$$ Note: The integration can be solved by basic integral formula $$\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.