Question

Find the following integral
$\int{\dfrac{1}{x\left( {{x}^{n}}-1 \right)}dx}$
[a] $\dfrac{1}{n}\ln \left| \dfrac{{{x}^{n}}}{{{x}^{n}}+1} \right|+C$
[b] $\dfrac{1}{n}\ln \left| \dfrac{{{x}^{n}}-1}{{{x}^{n}}} \right|+C$

Answer 1

Hint: Multiply numerator and denominator by ${{x}^{n-1}}$ and put $t={{x}^{n}}$. Use the fact that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ and hence prove that the given integral is equal to $\dfrac{1}{n}\int{\dfrac{1}{t\left( t-1 \right)}dt}$. Convert $\dfrac{1}{t\left( t-1 \right)}$ into partial fractions and use the fact that integral is linear, i.e. $\int{\left( f\left( x \right)+g\left( x \right) \right)dx}=\int{f\left( x \right)dx}+\int{g\left( x \right)dx}$. Hence evaluate the given integral. Verify your solution.

Complete step-by-step answer:
Let $I=\int{\dfrac{dx}{x\left( {{x}^{n}}-1 \right)}}$
Multiplying numerator and denominator of the integrand by ${{x}^{n-1}}$, we get
$I=\int{\dfrac{{{x}^{n-1}}dx}{{{x}^{n}}\left( {{x}^{n}}-1 \right)}}$
Put $t={{x}^{n}}$
Differentiating both sides with respect to x, we get
$\dfrac{dt}{dx}=n{{x}^{n-1}}$
Hence, we have $dt=n{{x}^{n-1}}dx$
Dividing both sides by n, we get
$\dfrac{dt}{n}={{x}^{n-1}}dx$
Hence, we have
$I=\int{\dfrac{dt}{nt\left( t-1 \right)}}$
We know that $\int{af\left( x \right)dx}=a\int{f\left( x \right)dx}$, where a is a scalar.
Hence, we have
$I=\dfrac{1}{n}\int{\dfrac{dt}{t\left( t-1 \right)}}$
We will convert $\dfrac{1}{t\left( t-1 \right)}$ into partial fractions.
So let $\dfrac{1}{t\left( t-1 \right)}=\dfrac{A}{t}+\dfrac{B}{t-1}$
Multiplying both sides by $t\left( t-1 \right)$, we get
$1=A\left( t-1 \right)+Bt$
Put t= 0, we get
$A\left( -1 \right)=1\Rightarrow A=-1$
Put t = 1, we get
$1=A\left( 1-1 \right)+B\Rightarrow B=1$
Hence, we have
$\dfrac{1}{t\left( t-1 \right)}=\dfrac{-1}{t}+\dfrac{1}{t-1}$
Hence the integral becomes
$I=\dfrac{1}{n}\int{\left( \dfrac{-1}{t}+\dfrac{1}{t-1} \right)dt}$
Using linearity of integration, i.e. $\int{\left( f\left( x \right)+g\left( x \right) \right)dx}=\int{f\left( x \right)dx}+\int{g\left( x \right)dx}$, we get
$I=\dfrac{-1}{n}\int{\dfrac{dt}{t}}+\dfrac{1}{n}\int{\dfrac{dt}{t-1}}$
We know that $\int{\dfrac{dx}{ax+b}}=\dfrac{1}{a}\ln \left| ax+b \right|+C$
Hence, we have
$I=-\dfrac{1}{n}\ln \left| t \right|+\dfrac{1}{n}\ln \left| t-1 \right|+C$, where C is the constant of integration.
Taking $\dfrac{1}{n}$ common, we get
$I=\dfrac{1}{n}\left( \ln \left| t-1 \right|-\ln \left| t \right| \right)+C$
We know that $\ln m-\ln n=\ln \dfrac{m}{n}$. Hence, we have
$I=\dfrac{1}{n}\ln \left| \dfrac{t-1}{t} \right|+C$
Reverting to the original variable, we get
$I=\dfrac{1}{n}\ln \left| \dfrac{{{x}^{n}}-1}{{{x}^{n}}} \right|+C$, which is the required result.
Hence option [b] is correct.

Note: Verification:
In questions where the integrand is not too complicated, it is often a good practice to verify the correctness of the solution by checking that the derivative of the integral is equal to the integrand.
In this case, we have
$I=\dfrac{1}{n}\ln \left| {{x}^{n}}-1 \right|-\dfrac{1}{n}\ln \left| {{x}^{n}} \right|+C$
Differentiating both sides, we get
$\dfrac{dI}{dx}=\dfrac{1}{n}\times \dfrac{1}{{{x}^{n}}-1}\times \dfrac{d}{dx}\left( {{x}^{n}}-1 \right)-\dfrac{1}{n}\times \dfrac{1}{{{x}^{n}}}\times \dfrac{d}{dx}\left( {{x}^{n}} \right)$
Hence, we have
$\begin{align}
  & \dfrac{dI}{dx}=\dfrac{1}{n}\times \dfrac{1}{{{x}^{n}}-1}\left( n{{x}^{n-1}} \right)-\dfrac{1}{n}\times \dfrac{1}{{{x}^{n}}}\times n{{x}^{n-1}} \\
 & =\dfrac{{{x}^{n-1}}}{{{x}^{n}}-1}-\dfrac{{{x}^{n-1}}}{{{x}^{n}}} \\
\end{align}$
Taking ${{x}^{n-1}}$ common, we get
$\dfrac{dI}{dx}={{x}^{n-1}}\times \left( \dfrac{1}{{{x}^{n}}-1}-\dfrac{1}{{{x}^{n}}} \right)=\dfrac{{{x}^{n-1}}}{{{x}^{n}}\left( {{x}^{n}}-1 \right)}\left( {{x}^{n}}-{{x}^{n}}+1 \right)=\dfrac{1}{x\left( {{x}^{n}}-1 \right)}$
Hence the derivative of the integral is equal to the integrand. Hence our solution is verified to be correct.