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Integrate ln10·log₁₀(1/A) with A in Powers of 2 | JEE

JEE Maths question with a full step-by-step solution.

Question
Let A=2 −log⁡10 ⁣100−xx / log⁡102A=2^{\,-\log_{10}\!\frac{100-x}{x}\,/\,\log_{10}2}. Then ∫ln⁡10⋅log⁡10 ⁣(1A)dx\displaystyle\int \ln 10\cdot\log_{10}\!\left(\dfrac{1}{A}\right)dx equals
A(x−100)ln⁡(100−x)−xln⁡x+C(x-100)\ln(100-x)-x\ln x+Ccorrect
B(x−100)ln⁡(100−x)+xln⁡x+C(x-100)\ln(100-x)+x\ln x+C
C(100−x)ln⁡(100−x)−xln⁡x+C(100-x)\ln(100-x)-x\ln x+C
D(100−x)ln⁡(100−x)+xln⁡x+C(100-x)\ln(100-x)+x\ln x+C
Solution
Step 1: Simplify AA. The exponent is −log⁡10100−xxlog⁡102=−log⁡2100−xx=log⁡2x100−x-\dfrac{\log_{10}\frac{100-x}{x}}{\log_{10}2}=-\log_{2}\dfrac{100-x}{x}=\log_{2}\dfrac{x}{100-x}, so
A=2log⁡2x100−x=x100−x  ⇒  1A=100−xxA=2^{\log_{2}\frac{x}{100-x}}=\dfrac{x}{100-x}\;\Rightarrow\;\dfrac{1}{A}=\dfrac{100-x}{x}
 Step 2: Simplify the integrand, using ln⁡10⋅log⁡10(⋅)=ln⁡(⋅)\ln 10\cdot\log_{10}(\cdot)=\ln(\cdot):
ln⁡10⋅log⁡10 ⁣(1A)=ln⁡100−xx=ln⁡(100−x)−ln⁡x\ln 10\cdot\log_{10}\!\left(\dfrac{1}{A}\right)=\ln\dfrac{100-x}{x}=\ln(100-x)-\ln x
 Step 3: Integrate ln⁡(100−x)\ln(100-x) by the substitution w=100−xw=100-x (dw=−dxdw=-dx):
∫ln⁡(100−x) dx=−∫ln⁡w dw=−(wln⁡w−w)=−(100−x)ln⁡(100−x)+(100−x)\int\ln(100-x)\,dx=-\int\ln w\,dw=-(w\ln w-w)=-(100-x)\ln(100-x)+(100-x)
 and ∫ln⁡x dx=xln⁡x−x\displaystyle\int\ln x\,dx=x\ln x-x. Step 4: Combine:
I=[−(100−x)ln⁡(100−x)+(100−x)]−[xln⁡x−x]+CI=\big[-(100-x)\ln(100-x)+(100-x)\big]-\big[x\ln x-x\big]+C
 The constants (100−x)+x=100(100-x)+x=100 are absorbed into CC:
I=(x−100)ln⁡(100−x)−xln⁡x+CI=(x-100)\ln(100-x)-x\ln x+C
 Correct answer: (1)
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