Question

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

Answer 1

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.