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Nested-Log Integral with ln(x·ln(x·lnx)) | JEE

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

Question
lnxln(xlnx)ln(xln(xlnx))+1+lnxlnxln(xlnx)dx\displaystyle\int \dfrac{\ln x\cdot \ln(x\ln x)\cdot \ln\big(x\ln(x\ln x)\big)+1+\ln x}{\ln x\cdot \ln(x\ln x)}\,dx equals
Axlnxxln(ln(xlnx))+x+Cx\ln x-x\ln(\ln(x\ln x))+x+C
Bxlnxxln(ln(xlnx))x+Cx\ln x-x\ln(\ln(x\ln x))-x+C
Cxlnx+xln(ln(xlnx))x+Cx\ln x+x\ln(\ln(x\ln x))-x+Ccorrect
Dxlnx+xln(ln(xlnx))+x+Cx\ln x+x\ln(\ln(x\ln x))+x+C
Solution
Step 1: Expand the triple-log factor with ln(ab)=lna+lnb\ln(ab)=\ln a+\ln b:
ln(xln(xlnx))=lnx+ln(ln(xlnx))\ln\big(x\ln(x\ln x)\big)=\ln x+\ln\big(\ln(x\ln x)\big)
Step 2: Divide the whole integrand by lnxln(xlnx)\ln x\cdot \ln(x\ln x). The first term loses its denominator:
integrand=[lnx+ln(ln(xlnx))]+1+lnxlnxln(xlnx)\text{integrand}=\Big[\ln x+\ln\big(\ln(x\ln x)\big)\Big]+\dfrac{1+\ln x}{\ln x\cdot \ln(x\ln x)}
Step 3: Identify a derivative. For h(x)=xln(ln(xlnx))h(x)=x\,\ln\big(\ln(x\ln x)\big), a direct differentiation gives
h(x)=ln(ln(xlnx))+1+lnxlnxln(xlnx)h'(x)=\ln\big(\ln(x\ln x)\big)+\dfrac{1+\ln x}{\ln x\cdot \ln(x\ln x)}
so the integrand is lnx+h(x)\ln x+h'(x). Step 4: Integrate. Using lnxdx=xlnxx\displaystyle\int\ln x\,dx=x\ln x-x,
I=(xlnxx)+xln(ln(xlnx))+C=xlnx+xln(ln(xlnx))x+CI=(x\ln x-x)+x\,\ln\big(\ln(x\ln x)\big)+C=x\ln x+x\ln(\ln(x\ln x))-x+C
Correct answer: (3)
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