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Find the value of the following integral
\[\int{\dfrac{1}{x{{\left( \log x \right)}^{n}}}dx}\]

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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.

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,
Here, by multiplying dt on both sides, we will get,
By cancelling the like terms from RHS, we will get,
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,
Therefore, we can write the integral as,
\[I=\int{\left( {{t}^{-n}} \right)dt}\]
Now, we know that
So we get the integral as,
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.

Last updated date: 21st Sep 2023
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