Answer
424.8k+ views
Hint: First multiply numerator and denominator by \[{{\text{x}}^{{\text{n - 1}}}}\]and assume \[{{\text{x}}^{\text{n}}}{\text{ = t}}\] and simplifying the problem. Then the integration will turn out to be a easier fraction to deal with.
Complete step by step solution: Given,
\[\int {\dfrac{{{\text{dx}}}}{{{\text{x(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}} \]
On Multiplying denominator and numerator by \[{{\text{x}}^{{\text{n - 1}}}}\],
\[ = \int {\dfrac{{{{\text{x}}^{{\text{n - 1}}}}{\text{dx}}}}{{{{\text{x}}^{\text{n}}}{\text{(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}} \]
Now, let, \[{{\text{x}}^{\text{n}}}{\text{ = t}}\]
\[ \Rightarrow {{\text{x}}^{{\text{n - 1}}}}{\text{dx = }}\dfrac{{{\text{dt}}}}{{\text{n}}}\],
So we get,
\[{\text{ = }}\dfrac{{\text{1}}}{{\text{n}}}\int {\dfrac{{{\text{dt}}}}{{{\text{t(t + 1)}}}}} \]………(1)
Now, we use the form of partial fractions to find the integral,
\[
\dfrac{1}{{t(t + 1)}} = \dfrac{A}{t} + \dfrac{B}{{t + 1}} \\
\Rightarrow \dfrac{1}{{t(t + 1)}} = \dfrac{{A(t + 1) + Bt}}{{t(t + 1)}} \\
\]
So,
\[A(t + 1) + Bt\]= 1
If we take, \[t = 0\] then,
\[A(0 + 1) + B.0 = 1\] gives us \[A = 1\]
And also for , \[t = - 1\]
\[A( - 1 + 1) + B. - 1 = 1\] gives us \[B = - 1\]
Now equation 1 can be written as,
\[ = \dfrac{1}{n}[\int {(\dfrac{1}{t} - \dfrac{1}{{t + 1}})dt} \]
If we integrate now, we have, as integral of \[\dfrac{1}{t}\] is \[\log t\],
\[ \Rightarrow \dfrac{1}{n}[\log t - \log (t + 1)]\]
Now, also,
`\[\log a - \log b = \log \dfrac{a}{b}\]
\[ \Rightarrow \dfrac{1}{n}[\log (\dfrac{t}{{t + 1}})]\]
Putting the value of t we have,
\[ = \dfrac{1}{n}[\log (\dfrac{{{x^n}}}{{{x^n} + 1}})]\]
So, we get, \[\int {\dfrac{{{\text{dx}}}}{{{\text{x(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}} \]\[ = \dfrac{1}{n}[\log (\dfrac{{{x^n}}}{{{x^n} + 1}})]+C\]
Note: \[\dfrac{1}{{t(t + 1)}}\]can be directly written as sum of two fractions \[\dfrac{1}{t}, - \dfrac{1}{{t + 1}}\]as \[\dfrac{1}{{t(t + 1)}} = \dfrac{1}{t} - \dfrac{1}{{t + 1}}\]. We can also do this integration in another way by multiplying the denominator and numerator by \[{x^n} - 1\].
Complete step by step solution: Given,
\[\int {\dfrac{{{\text{dx}}}}{{{\text{x(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}} \]
On Multiplying denominator and numerator by \[{{\text{x}}^{{\text{n - 1}}}}\],
\[ = \int {\dfrac{{{{\text{x}}^{{\text{n - 1}}}}{\text{dx}}}}{{{{\text{x}}^{\text{n}}}{\text{(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}} \]
Now, let, \[{{\text{x}}^{\text{n}}}{\text{ = t}}\]
\[ \Rightarrow {{\text{x}}^{{\text{n - 1}}}}{\text{dx = }}\dfrac{{{\text{dt}}}}{{\text{n}}}\],
So we get,
\[{\text{ = }}\dfrac{{\text{1}}}{{\text{n}}}\int {\dfrac{{{\text{dt}}}}{{{\text{t(t + 1)}}}}} \]………(1)
Now, we use the form of partial fractions to find the integral,
\[
\dfrac{1}{{t(t + 1)}} = \dfrac{A}{t} + \dfrac{B}{{t + 1}} \\
\Rightarrow \dfrac{1}{{t(t + 1)}} = \dfrac{{A(t + 1) + Bt}}{{t(t + 1)}} \\
\]
So,
\[A(t + 1) + Bt\]= 1
If we take, \[t = 0\] then,
\[A(0 + 1) + B.0 = 1\] gives us \[A = 1\]
And also for , \[t = - 1\]
\[A( - 1 + 1) + B. - 1 = 1\] gives us \[B = - 1\]
Now equation 1 can be written as,
\[ = \dfrac{1}{n}[\int {(\dfrac{1}{t} - \dfrac{1}{{t + 1}})dt} \]
If we integrate now, we have, as integral of \[\dfrac{1}{t}\] is \[\log t\],
\[ \Rightarrow \dfrac{1}{n}[\log t - \log (t + 1)]\]
Now, also,
`\[\log a - \log b = \log \dfrac{a}{b}\]
\[ \Rightarrow \dfrac{1}{n}[\log (\dfrac{t}{{t + 1}})]\]
Putting the value of t we have,
\[ = \dfrac{1}{n}[\log (\dfrac{{{x^n}}}{{{x^n} + 1}})]\]
So, we get, \[\int {\dfrac{{{\text{dx}}}}{{{\text{x(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}} \]\[ = \dfrac{1}{n}[\log (\dfrac{{{x^n}}}{{{x^n} + 1}})]+C\]
Note: \[\dfrac{1}{{t(t + 1)}}\]can be directly written as sum of two fractions \[\dfrac{1}{t}, - \dfrac{1}{{t + 1}}\]as \[\dfrac{1}{{t(t + 1)}} = \dfrac{1}{t} - \dfrac{1}{{t + 1}}\]. We can also do this integration in another way by multiplying the denominator and numerator by \[{x^n} - 1\].
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)