Answer

Verified

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.

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.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Assertion The resistivity of a semiconductor increases class 13 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

Using the following information to help you answer class 12 chemistry CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Two charges are placed at a certain distance apart class 12 physics CBSE

Difference Between Plant Cell and Animal Cell

What organs are located on the left side of your body class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

The planet nearest to earth is A Mercury B Venus C class 6 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is BLO What is the full form of BLO class 8 social science CBSE