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The value of \[\int {\dfrac{{{\text{dx}}}}{{{\text{x(}}{{\text{x}}^{\text{n}}}{\text{ + 1)}}}}} \]is
a. \[\dfrac{{\text{1}}}{{\text{n}}}{\text{log(}}\dfrac{{{{\text{x}}^{\text{n}}}}}{{{{\text{x}}^{\text{n}}}{\text{ + 1}}}}{\text{) + C}}\]
b. \[{\text{log(}}\dfrac{{{{\text{x}}^{\text{n}}} + 1}}{{{{\text{x}}^{\text{n}}}}}{\text{) + C}}\]
c. \[\dfrac{{\text{1}}}{{\text{n}}}{\text{log(}}\dfrac{{{{\text{x}}^{\text{n}}}{\text{ + 1}}}}{{{{\text{x}}^{\text{n}}}}}{\text{) + C}}\]
d. \[{\text{log(}}\dfrac{{{{\text{x}}^{\text{n}}}}}{{{{\text{x}}^{\text{n}}} + 1}}{\text{) + C}}\]

Answer
VerifiedVerified
129.3k+ 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 \[{x^n} - 1\].