Question

integral of 1/logx
pls answer fast​

Answer 1

$\large \: \red{\mathcal{ANSWER}} \\ \:$

$\small{\Longrightarrow \: I \: = \int \: \frac{1}{log_{a}(x)} \: dx }\: \: \small\\ \Longrightarrow \: let \: \: \: \: \small \: log(x) = t \\ \Longrightarrow \: then \: \: \: \: \: x = {e}^{t} \\\Longrightarrow \: or \: dx \: = {e}^{t} \: dt$

$\small \: \Longrightarrow \: I \: = \: \int \frac{ {e}^{t} }{t} dt \\ \small\Longrightarrow I \: = \: \int \frac{1 + \frac{t}{1! } + \frac{ {t}^{2} }{2!} + ............... }{t}$

$\small \: \Longrightarrow \: I \: = \int \frac{1}{t} + \frac{1}{1} + \frac{t}{2! } + .........dt$

$\blue{\small \: \Longrightarrow \: I \: = \: log(t) + t + \frac{ {t}^{2} }{4} + \frac{ {t}^{3} }{2 \times 3! } + c........}$

where c is constant of integration

$\mathrm{ \small\Longrightarrow \: series \: expansion \: of \: } { {e}^{t} }\\ \small\Longrightarrow \boxed{ {e}^{t} \: = \: 1 + \frac{t}{1! } + \frac{ {t}^{2} }{2!} + \frac{ {t}^{3} }{3!} ............... \: }$