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Match List-I with List-II

(Consider log⁡ex=log⁡x\log_e x = \log x)

List-IList-II
IntegralSolution: where C is an arbitrary constant
(A) ∫f′(x)f(x)log⁡f(x) dx\displaystyle\int \dfrac{f'(x)}{f(x)\log f(x)}\,dx(I) xlog⁡(log⁡x)+cx\log(\log x) + c
(B) ∫(log⁡(log⁡x)+1log⁡x)dx\displaystyle\int \left(\log(\log x) + \dfrac{1}{\log x}\right) dx(II) xlog⁡x+C\dfrac{x}{\log x} + C
(C) ∫log⁡x(1+log⁡x)2 dx\displaystyle\int \dfrac{\log x}{(1+\log x)^2}\,dx(III) log⁡∣log⁡(f(x))∣+c\log\left\vert \log(f(x)) \right\vert + c
(D) ∫(1log⁡x−1(log⁡x)2)dx\displaystyle\int \left(\dfrac{1}{\log x} - \dfrac{1}{(\log x)^2}\right) dx(IV) x1+log⁡x+c\dfrac{x}{1+\log x} + c

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

In (A), substituting u=log⁡f(x)u=\log f(x) gives log⁡∣log⁡f(x)∣\log|\log f(x)|, so (A)-(III).

Differentiating the List-II answers confirms the rest: ddx[xlog⁡(log⁡x)]=log⁡(log⁡x)+1log⁡x\frac{d}{dx}[x\log(\log x)]=\log(\log x)+\frac{1}{\log x} gives (B)-(I); ddx[x1+log⁡x]=log⁡x(1+log⁡x)2\frac{d}{dx}\left[\frac{x}{1+\log x}\right]=\frac{\log x}{(1+\log x)^2} gives (C)-(IV); ddx[xlog⁡x]=1log⁡x−1(log⁡x)2\frac{d}{dx}\left[\frac{x}{\log x}\right]=\frac{1}{\log x}-\frac{1}{(\log x)^2} gives (D)-(II).

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