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

Answer

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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.

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.

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