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Hint: Here, to find the value of the given integral, first of all take log x = t and then substitute all the variables of x that are \[\dfrac{dx}{x}\] and \[\log x\] in terms of t in the given integral.

Here we have to find the value of \[\int{\dfrac{1}{x{{\left( \log x \right)}^{n}}}dx}\].

Let us consider the given integral as

\[I=\int{\dfrac{1.dx}{x{{\left( \log x \right)}^{n}}}}....\left( i \right)\]

Now, as we can see that this integral contains both \[\dfrac{1}{x}\] and \[\log x\], therefore let us consider log x = t

Since, we know that \[\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}\], therefore by differentiating both sides, we get,

\[\dfrac{1dx}{xdt}=1\]

Here, by multiplying dt on both sides, we will get,

\[\dfrac{1dx}{xdt}.dt=1.dt\]

By cancelling the like terms from RHS, we will get,

\[\dfrac{dx}{x}=dt\]

Now, we will put the values of \[\dfrac{dx}{x}\text{ and }\log x\]in terms of t in equation (i). We get

\[I=\int{\dfrac{dt}{{{\left( t \right)}^{n}}}}\]

Since, we know that,

\[\dfrac{1}{{{a}^{n}}}={{a}^{-n}}\]

Therefore, we can write the integral as,

\[I=\int{\left( {{t}^{-n}} \right)dt}\]

Now, we know that

\[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+k}\]

So we get the integral as,

\[I=\dfrac{{{t}^{-n+1}}}{-n+1}+k\]

We know that we should always convert the

So, here as we had assumed that log x = t, so now, we will replace “t” in terms of “x”. So, we will get the integral as

\[I=\dfrac{{{\left( \log x \right)}^{-n+1}}}{\left( -n+1 \right)}+k\]

Therefore, we finally get the value of integral as,

\[I=\int{\dfrac{1.dx}{x{{\left( \log x \right)}^{n}}}=\dfrac{{{\left( \log x \right)}^{1-n}}}{\left( 1-n \right)}}+k\]

Note: Whenever \[\dfrac{1}{x}\text{ and }\log x\] come together in question, students should always use this approach of putting log x = t and then differentiating it to get \[\dfrac{1}{x}dx\] which makes the solution feasible. Also, students should always remember to convert the assumed variable into the original variable at the end of the solution, in this question, it is t into x.

__Complete step-by-step answer:__Here we have to find the value of \[\int{\dfrac{1}{x{{\left( \log x \right)}^{n}}}dx}\].

Let us consider the given integral as

\[I=\int{\dfrac{1.dx}{x{{\left( \log x \right)}^{n}}}}....\left( i \right)\]

Now, as we can see that this integral contains both \[\dfrac{1}{x}\] and \[\log x\], therefore let us consider log x = t

Since, we know that \[\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}\], therefore by differentiating both sides, we get,

\[\dfrac{1dx}{xdt}=1\]

Here, by multiplying dt on both sides, we will get,

\[\dfrac{1dx}{xdt}.dt=1.dt\]

By cancelling the like terms from RHS, we will get,

\[\dfrac{dx}{x}=dt\]

Now, we will put the values of \[\dfrac{dx}{x}\text{ and }\log x\]in terms of t in equation (i). We get

\[I=\int{\dfrac{dt}{{{\left( t \right)}^{n}}}}\]

Since, we know that,

\[\dfrac{1}{{{a}^{n}}}={{a}^{-n}}\]

Therefore, we can write the integral as,

\[I=\int{\left( {{t}^{-n}} \right)dt}\]

Now, we know that

\[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+k}\]

So we get the integral as,

\[I=\dfrac{{{t}^{-n+1}}}{-n+1}+k\]

We know that we should always convert the

So, here as we had assumed that log x = t, so now, we will replace “t” in terms of “x”. So, we will get the integral as

\[I=\dfrac{{{\left( \log x \right)}^{-n+1}}}{\left( -n+1 \right)}+k\]

Therefore, we finally get the value of integral as,

\[I=\int{\dfrac{1.dx}{x{{\left( \log x \right)}^{n}}}=\dfrac{{{\left( \log x \right)}^{1-n}}}{\left( 1-n \right)}}+k\]

Note: Whenever \[\dfrac{1}{x}\text{ and }\log x\] come together in question, students should always use this approach of putting log x = t and then differentiating it to get \[\dfrac{1}{x}dx\] which makes the solution feasible. Also, students should always remember to convert the assumed variable into the original variable at the end of the solution, in this question, it is t into x.

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