Question

Find the value of \[\int {\dfrac{1}{{x\left( {{x^n} + 1} \right)}}dx} \].
A.\[\dfrac{1}{n}\log \left( {\dfrac{{{x^n}}}{{{x^n} + 1}}} \right) + C\]
B.\[\log \left( {\dfrac{{{x^n} + 1}}{{{x^n}}}} \right) + C\]
C.\[\dfrac{1}{n}\log \left( {\dfrac{{{x^n} + 1}}{{{x^n}}}} \right) + C\]
D.\[\log \left( {\dfrac{{{x^n}}}{{{x^n} + 1}}} \right) + C\]

Answer 1

Hint: Here, we will take \[u = {x^n} + 1\] and then differentiate it with respect to \[x\]. Then we will take the partial fraction of \[\dfrac{1}{{u\left( {u - 1} \right)}}\] and then use the integral properties to find the required value.

Complete step-by-step answer:
We are given that
\[\int {\dfrac{1}{{x\left( {{x^n} + 1} \right)}}dx} \]
Taking \[u = {x^n} + 1\] and then differentiate it with respect to \[x\], we get
\[ \Rightarrow \dfrac{{du}}{{dx}} = n{x^{n - 1}}\]
Cross-multiplying the above equation, we get
\[ \Rightarrow du = n{x^{n - 1}}dx\]
Substituting the value of \[u\] and \[du\] in the given equation, we get
\[
   \Rightarrow \int {\dfrac{1}{{nu\left( {u - 1} \right)}}du} \\
   \Rightarrow \dfrac{1}{n}\int {\dfrac{1}{{u\left( {u - 1} \right)}}du} \\
 \]
Taking the partial fraction of \[\dfrac{1}{{u\left( {u - 1} \right)}}\] in the above equation, we get
\[
   \Rightarrow \dfrac{1}{n}\int {\left( { - \dfrac{1}{u} + \dfrac{1}{{u - 1}}} \right)du} \\
   \Rightarrow - \dfrac{1}{n}\int {\dfrac{1}{u}du} + \dfrac{1}{n}\int {\dfrac{1}{{u - 1}}du} \\
 \]
Using the value \[\int {\dfrac{1}{u}du} = \ln \left| u \right| + C\] in the above equation, we get
\[
   \Rightarrow - \dfrac{1}{n}\ln \left| u \right| + \dfrac{1}{n}\ln \left| {u - 1} \right| + C \\
   \Rightarrow \dfrac{1}{n}\left( { - \ln \left| u \right| + \ln \left| {u - 1} \right|} \right) + C \\
   \Rightarrow \dfrac{1}{n}\left( {\ln \left| {u - 1} \right| - \ln \left| u \right|} \right) + C \\
 \]
Using the logarithm value \[\ln a - \ln b = \ln \dfrac{a}{b}\] in the above equation, we get
\[ \Rightarrow \dfrac{1}{n}\ln \left( {\dfrac{{u - 1}}{u}} \right)\]

Substituting \[u = {x^n} + 1\] back in the above equation, we get
\[
   \Rightarrow \dfrac{1}{n}\ln \left( {\dfrac{{{x^n} + 1 - 1}}{{{x^n} + 1}}} \right) + C \\
   \Rightarrow \dfrac{1}{n}\ln \left( {\dfrac{{{x^n}}}{{{x^n} + 1}}} \right) + C \\
 \]
Hence, option A is correct.

Note: We need to know that while finding the value of indefinite integral, we have to add the constant in the final answer or else the answer will be incomplete. We have to be really thorough with the integrations and differentiation of the functions. The key point in this question is to use the integration by partial sums and integration rule,\[\int {\dfrac{1}{u}du} = \ln \left| u \right| + C\] to solve this problem. Do not forget that many integrals can be evaluated in multiple ways and so more than one technique may be used on it, but this problem can only be solved by parts.