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Integral with (ln(x+1)−lnx)¹¹ in Denominator | JEE

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Question
∫dxx(x+1) (ln⁡(x+1)−ln⁡x)11\displaystyle\int \dfrac{dx}{x(x+1)\,\big(\ln(x+1)-\ln x\big)^{11}} equals
A110(ln⁡(x+1)−ln⁡x)10+C\dfrac{1}{10\big(\ln(x+1)-\ln x\big)^{10}}+Ccorrect
B(ln⁡(x+1)−ln⁡x)1010+C\dfrac{\big(\ln(x+1)-\ln x\big)^{10}}{10}+C
C111(ln⁡(x+1)−ln⁡x)11+C\dfrac{1}{11\big(\ln(x+1)-\ln x\big)^{11}}+C
D(ln⁡(x+1)−ln⁡x)1111+C\dfrac{\big(\ln(x+1)-\ln x\big)^{11}}{11}+C
Solution
Step 1: Substitute t=ln⁡(x+1)−ln⁡x=ln⁡x+1xt=\ln(x+1)-\ln x=\ln\dfrac{x+1}{x}. Step 2: Differentiate:
dt=(1x+1−1x)dx=x−(x+1)x(x+1) dx=−dxx(x+1)dt=\left(\dfrac{1}{x+1}-\dfrac{1}{x}\right)dx=\dfrac{x-(x+1)}{x(x+1)}\,dx=-\dfrac{dx}{x(x+1)}
⇒dxx(x+1)=− dt\Rightarrow \dfrac{dx}{x(x+1)}=-\,dt
 Step 3: Rewrite the integral:
I=∫1t11 (−dt)=−∫t−11 dtI=\int\dfrac{1}{t^{11}}\,(-dt)=-\int t^{-11}\,dt
 Step 4: Integrate:
=−t−10−10=t−1010=110 t10=-\dfrac{t^{-10}}{-10}=\dfrac{t^{-10}}{10}=\dfrac{1}{10\,t^{10}}
 Step 5: Back-substitute t=ln⁡(x+1)−ln⁡xt=\ln(x+1)-\ln x:
I=110(ln⁡(x+1)−ln⁡x)10+CI=\dfrac{1}{10\big(\ln(x+1)-\ln x\big)^{10}}+C
 Correct answer: (1)
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