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Question: Find the following integral \(\int{\dfrac{1}{x\left( {{x}^{n}}-1 \right)}dx}\) [a] \(\dfrac{1}{...

Find the following integral
∫1x(xn−1)dx\int{\dfrac{1}{x\left( {{x}^{n}}-1 \right)}dx}
[a] 1nln⁡∣xnxn+1∣+C\dfrac{1}{n}\ln \left| \dfrac{{{x}^{n}}}{{{x}^{n}}+1} \right|+C
[b] 1nln⁡∣xn−1xn∣+C\dfrac{1}{n}\ln \left| \dfrac{{{x}^{n}}-1}{{{x}^{n}}} \right|+C

Explanation

Solution

Hint: Multiply numerator and denominator by xn−1{{x}^{n-1}} and put t=xnt={{x}^{n}}. Use the fact that ddx(xn)=nxn−1\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}} and hence prove that the given integral is equal to 1n∫1t(t−1)dt\dfrac{1}{n}\int{\dfrac{1}{t\left( t-1 \right)}dt}. Convert 1t(t−1)\dfrac{1}{t\left( t-1 \right)} into partial fractions and use the fact that integral is linear, i.e. ∫(f(x)+g(x))dx=∫f(x)dx+∫g(x)dx\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=∫dxx(xn−1)I=\int{\dfrac{dx}{x\left( {{x}^{n}}-1 \right)}}
Multiplying numerator and denominator of the integrand by xn−1{{x}^{n-1}}, we get
I=∫xn−1dxxn(xn−1)I=\int{\dfrac{{{x}^{n-1}}dx}{{{x}^{n}}\left( {{x}^{n}}-1 \right)}}
Put t=xnt={{x}^{n}}
Differentiating both sides with respect to x, we get
dtdx=nxn−1\dfrac{dt}{dx}=n{{x}^{n-1}}
Hence, we have dt=nxn−1dxdt=n{{x}^{n-1}}dx
Dividing both sides by n, we get
dtn=xn−1dx\dfrac{dt}{n}={{x}^{n-1}}dx
Hence, we have
I=∫dtnt(t−1)I=\int{\dfrac{dt}{nt\left( t-1 \right)}}
We know that ∫af(x)dx=a∫f(x)dx\int{af\left( x \right)dx}=a\int{f\left( x \right)dx}, where a is a scalar.
Hence, we have
I=1n∫dtt(t−1)I=\dfrac{1}{n}\int{\dfrac{dt}{t\left( t-1 \right)}}
We will convert 1t(t−1)\dfrac{1}{t\left( t-1 \right)} into partial fractions.
So let 1t(t−1)=At+Bt−1\dfrac{1}{t\left( t-1 \right)}=\dfrac{A}{t}+\dfrac{B}{t-1}
Multiplying both sides by t(t−1)t\left( t-1 \right), we get
1=A(t−1)+Bt1=A\left( t-1 \right)+Bt
Put t= 0, we get
A(−1)=1⇒A=−1A\left( -1 \right)=1\Rightarrow A=-1
Put t = 1, we get
1=A(1−1)+B⇒B=11=A\left( 1-1 \right)+B\Rightarrow B=1
Hence, we have
1t(t−1)=−1t+1t−1\dfrac{1}{t\left( t-1 \right)}=\dfrac{-1}{t}+\dfrac{1}{t-1}
Hence the integral becomes
I=1n∫(−1t+1t−1)dtI=\dfrac{1}{n}\int{\left( \dfrac{-1}{t}+\dfrac{1}{t-1} \right)dt}
Using linearity of integration, i.e. ∫(f(x)+g(x))dx=∫f(x)dx+∫g(x)dx\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=−1n∫dtt+1n∫dtt−1I=\dfrac{-1}{n}\int{\dfrac{dt}{t}}+\dfrac{1}{n}\int{\dfrac{dt}{t-1}}
We know that ∫dxax+b=1aln⁡∣ax+b∣+C\int{\dfrac{dx}{ax+b}}=\dfrac{1}{a}\ln \left| ax+b \right|+C
Hence, we have
I=−1nln⁡∣t∣+1nln⁡∣t−1∣+CI=-\dfrac{1}{n}\ln \left| t \right|+\dfrac{1}{n}\ln \left| t-1 \right|+C, where C is the constant of integration.
Taking 1n\dfrac{1}{n} common, we get
I=1n(ln⁡∣t−1∣−ln⁡∣t∣)+CI=\dfrac{1}{n}\left( \ln \left| t-1 \right|-\ln \left| t \right| \right)+C
We know that ln⁡m−ln⁡n=ln⁡mn\ln m-\ln n=\ln \dfrac{m}{n}. Hence, we have
I=1nln⁡∣t−1t∣+CI=\dfrac{1}{n}\ln \left| \dfrac{t-1}{t} \right|+C
Reverting to the original variable, we get
I=1nln⁡∣xn−1xn∣+CI=\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=1nln⁡∣xn−1∣−1nln⁡∣xn∣+CI=\dfrac{1}{n}\ln \left| {{x}^{n}}-1 \right|-\dfrac{1}{n}\ln \left| {{x}^{n}} \right|+C
Differentiating both sides, we get
dIdx=1n×1xn−1×ddx(xn−1)−1n×1xn×ddx(xn)\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
dIdx=1n×1xn−1(nxn−1)−1n×1xn×nxn−1 =xn−1xn−1−xn−1xn \begin{aligned} & \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{aligned}
Taking xn−1{{x}^{n-1}} common, we get
dIdx=xn−1×(1xn−1−1xn)=xn−1xn(xn−1)(xn−xn+1)=1x(xn−1)\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.