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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

\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

x\ln x-x\ln(\ln(x\ln x))+x+C

x\ln x-x\ln(\ln(x\ln x))-x+C

x\ln x+x\ln(\ln(x\ln x))-x+C

correct

x\ln x+x\ln(\ln(x\ln x))+x+C

Solution

Step 1: Expand the triple-log factor with

\ln(ab)=\ln a+\ln b

\ln\big(x\ln(x\ln x)\big)=\ln x+\ln\big(\ln(x\ln x)\big)

Step 2: Divide the whole integrand by

\ln x\cdot \ln(x\ln x)

. The first term loses its denominator:

\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)=x\,\ln\big(\ln(x\ln x)\big)

, a direct differentiation gives

h'(x)=\ln\big(\ln(x\ln x)\big)+\dfrac{1+\ln x}{\ln x\cdot \ln(x\ln x)}

so the integrand is

\ln x+h'(x)

. Step 4: Integrate. Using

\displaystyle\int\ln x\,dx=x\ln x-x

I=(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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