Indefinite IntegrationeasyFree

Integral of 1/(x·ℓ(x)·ℓ²(x)···ℓⁿ(x)) | JEE

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

Question

f(x)=\dfrac{1}{x\cdot \ell(x)\cdot \ell^{2}(x)\cdot \ell^{3}(x)\cdots \ell^{n}(x)}

, where

\ell^{r}(x)

denotes

\log

applied

r

times, then

\displaystyle\int f(x)\,dx

equals

\ell^{n}(x)+C

\ell^{n+1}(x)+C

correct

\ell^{n-1}(x)+C

DNone of these

Solution

Step 1: Fix the notation:

\ell^{1}(x)=\ln x,\ \ell^{2}(x)=\ln(\ln x),\ \ldots,\ \ell^{n}(x)=\ln

applied

n

times. Step 2: Differentiate the candidate

\ell^{n+1}(x)

by the chain rule:

\dfrac{d}{dx}\,\ell^{n+1}(x)=\dfrac{1}{\ell^{n}(x)}\cdot\dfrac{d}{dx}\,\ell^{n}(x)

and in turn

\dfrac{d}{dx}\,\ell^{n}(x)=\dfrac{1}{\ell^{n-1}(x)}\cdot\dfrac{d}{dx}\,\ell^{n-1}(x)

, and so on down to

\dfrac{d}{dx}\,\ell^{1}(x)=\dfrac{1}{x}

. Step 3: Multiplying all

n+1

reciprocal factors together:

\dfrac{d}{dx}\,\ell^{n+1}(x)=\dfrac{1}{x\cdot \ell(x)\cdot \ell^{2}(x)\cdots \ell^{n}(x)}=f(x)

Step 4: Therefore

\displaystyle\int f(x)\,dx=\ell^{n+1}(x)+C

. Correct answer: (2)

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