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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=2log10 ⁣100xx/log102A=2^{\,-\log_{10}\!\frac{100-x}{x}\,/\,\log_{10}2}. Then ln10log10 ⁣(1A)dx\displaystyle\int \ln 10\cdot\log_{10}\!\left(\dfrac{1}{A}\right)dx equals
A(x100)ln(100x)xlnx+C(x-100)\ln(100-x)-x\ln x+Ccorrect
B(x100)ln(100x)+xlnx+C(x-100)\ln(100-x)+x\ln x+C
C(100x)ln(100x)xlnx+C(100-x)\ln(100-x)-x\ln x+C
D(100x)ln(100x)+xlnx+C(100-x)\ln(100-x)+x\ln x+C
Solution
Step 1: Simplify AA. The exponent is log10100xxlog102=log2100xx=log2x100x-\dfrac{\log_{10}\frac{100-x}{x}}{\log_{10}2}=-\log_{2}\dfrac{100-x}{x}=\log_{2}\dfrac{x}{100-x}, so
A=2log2x100x=x100x    1A=100xxA=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 ln10log10()=ln()\ln 10\cdot\log_{10}(\cdot)=\ln(\cdot):
ln10log10 ⁣(1A)=ln100xx=ln(100x)lnx\ln 10\cdot\log_{10}\!\left(\dfrac{1}{A}\right)=\ln\dfrac{100-x}{x}=\ln(100-x)-\ln x
Step 3: Integrate ln(100x)\ln(100-x) by the substitution w=100xw=100-x (dw=dxdw=-dx):
ln(100x)dx=lnwdw=(wlnww)=(100x)ln(100x)+(100x)\int\ln(100-x)\,dx=-\int\ln w\,dw=-(w\ln w-w)=-(100-x)\ln(100-x)+(100-x)
and lnxdx=xlnxx\displaystyle\int\ln x\,dx=x\ln x-x. Step 4: Combine:
I=[(100x)ln(100x)+(100x)][xlnxx]+CI=\big[-(100-x)\ln(100-x)+(100-x)\big]-\big[x\ln x-x\big]+C
The constants (100x)+x=100(100-x)+x=100 are absorbed into CC:
I=(x100)ln(100x)xlnx+CI=(x-100)\ln(100-x)-x\ln x+C
Correct answer: (1)
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