$\int \frac{1}{x log x log (log x)} d x$ = plog(log(logx))+C then p =

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Solution

$\int \frac{1}{x log x log (log x)} d x$
Put $y = log (log x) ⟹ d y = \frac{d x}{x log x}$
$\int \frac{d y}{y} = log y + C = log (log (log x)) + C$

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n

p

+

n \geq p + 2

D (n, p) ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^{n} C_{p} & ^{n} C_{p + 1} & ^{n} C_{p + 2}^{n + 1} C_{p} & ^{n + 1} C_{p + 1} & ^{n + 1} C_{p + 2}^{n + 2} C_{p} & ^{n + 3} C_{p + 1} & ^{n + 2} C_{p + 2} \end{matrix} ∣ ∣ ∣ ∣ ∣

D (n, p) = \frac{^{n + 2} C_{3}}{^{p + 2} C_{3}} D (n - 1, p - 1)

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