If lr(x) means log log log..... (x), the 'log' being repeated 'r' times, then ∫1x ℓ(x) ℓ2(x) ℓ3(x)−−−ℓr(x) dx=

t=lr+1 dt=1lr(x).........l(x).x ∫1 dt = t+c=lr+1(x)+c

If lr(x) means log log log..... (x), the 'log' being repeated 'r' times, then ∫1x ℓ(x) ℓ2(x) ℓ3(x)−−−ℓr(x) dx=

t=lr+1 dt=1lr(x).........l(x).x ∫1 dt = t+c=lr+1(x)+c

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If $l^{r} (x)$ means $\log \log \log ..... (x),$ the 'log' being repeated 'r' times, then
$\int \frac{1}{x ℓ (x) ℓ^{2} (x) ℓ^{3} (x) - - - ℓ^{r} (x)} d x =$

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$\frac{ℓ^{r + 1} (x)}{r + 1}$

$ℓ^{r + 1} (x) + c$

$\frac{ℓ^{r} (x)}{r} + c$

$ℓ^{r} (x) + c$

answer is B.

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Detailed Solution

$t = l^{r + 1}$

$d t = \frac{1}{l^{r} (x) ......... l (x) . x}$

$\int 1 d t = t + c = l^{r + 1} (x) + c$

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If lr(x) means log log log..... (x), the 'log' being repeated 'r' times, then ∫1x ℓ(x) ℓ2(x) ℓ3(x)−−−ℓr(x) dx=

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If $l^{r} (x)$ means $\log \log \log ..... (x),$ the 'log' being repeated 'r' times, then
$\int \frac{1}{x ℓ (x) ℓ^{2} (x) ℓ^{3} (x) - - - ℓ^{r} (x)} d x =$