Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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

seo-qna
Last updated date: 23rd Apr 2024
Total views: 424.5k
Views today: 8.24k
Answer
VerifiedVerified
424.5k+ views
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,
\[\dfrac{1dx}{xdt}=1\]
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.