i = ∫ log (logx) +1/logx^2dx ⇒ log(logx)dx+∫1/logx^2dx ⇒∫ 1.log(log x)dx #160; ⇒ #160; x ln (ln(x))∫1ln(x)dx ⇒ x ln (ln(x)) li(x)+c ...

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Monotonically Increasing Functions

∫loglogx+1/lo...

Question

$\int log (log x) + \frac{1}{{log x}^{2}} d x$

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Solution

$i = \int l o g (l o g x) + \frac{1}{l o g x^{2}} d x$
$\Rightarrow l o g (l o g x) d x + \int \frac{1}{l o g x^{2}} d x$
$\Rightarrow \int 1. l o g (l o g x) d x$
$\Rightarrow x l n (l n (x)) \int \frac{1}{l n (x)} d x$
$\Rightarrow x l n (l n (x)) - l i (x) + c$
${$ logarithmic intergel $}$
(II) $\int \frac{1}{l o g x^{2}} d x$
$\Rightarrow \frac{1}{2} \int \frac{1}{l o g x} d x$
$\Rightarrow + \frac{1}{2} l (x)$
(I) + (II)
$\Rightarrow x l n (l n (x)) - l i (x) + \frac{l i (x)}{2} + c$
$i \Rightarrow x l n (l n (x)) - \frac{l i (x)}{2} + c$

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