∫(1/(lnx))dx Help out


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 183687 by Mastermind last updated on 28/Dec/22

$\int \frac{1}{lnx} dx$ $Help out$
	Answered by a.lgnaoui last updated on 28/Dec/22

	$\frac{1}{\ln (x)} = \frac{xln (x)}{x {[\ln (x)]}^{2}} = (\frac{lnx}{x}) (\frac{x}{{[\ln (x)]}^{2}})$ $= d (lnx) lnx \frac{x}{{[\ln (x)]}^{2}} = \frac{xd [{(lnx)}^{2}]}{2 {[\ln (x)]}^{2}}$ $p o s o n s u = {[\ln (x)]}^{2} v = \frac{x}{{[\ln (x)]}^{2}}$ $V^{'} = \frac{{[\ln (x)]}^{2} - 2 lnx}{{[\ln (x)]}^{2}}$ $I = \frac{1}{2} [(x) - \int [\ln {(x)}^{2}] - 2 \ln (x)] dx]$ $= \frac{1}{2} [x - \int [{(lnx - 1)}^{2} - 1] dx]$ $= x - \frac{1}{2} \int {(lnx - 1)}^{2} dx$ $posons$ $P = {(lnx - 1)}^{2} dP = \frac{2}{x} (lnx - 1)$ $dQ = 1 Q = x$ $\int {(lnx - 1)}^{2} dx = x {(lnx - 1)}^{2} - 2 \int (lnx - 1) dx$ $= x {(lnx - 1)}^{2} + 4 x - 2 xlnx + c$ $soit :$ $I = x - \frac{1}{2} [x {(lnx - 1)}^{2} - 2 xlnx + 4 x + C$ $C o n c l u s i o n :$ $\int \frac{dx}{lnx} = x [lnx - \frac{1}{2} {(lnx - 1)}^{2} - 1] + C$ $j$
	Commented by Frix last updated on 28/Dec/22

	$Sorry but this is nonsense .$
	Answered by Frix last updated on 28/Dec/22

	$\int \frac{d x}{\ln x} = li x + C$ $[by definition : li x = Integral Logarithm]$
	Answered by paul2222 last updated on 28/Dec/22

	$let u = \ln (x)$ $x = e^{u}$ $dx = e^{u} du$ $\int \frac{e^{u}}{u} du$ $we recall e^{u} = \underset{n = 0}{\sum^{\infty}} \frac{u^{n}}{n!}$ $\underset{n = 0}{\sum^{\infty}} \frac{1}{n!} \int u^{n} du = \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} [\frac{u^{n + 1}}{n + 1}] + C$ $where u = \ln (x)$ $\underset{n = 0}{\sum^{\infty}} \frac{\ln {(x)}^{n + 1}}{n! (n + 1)} = \ln (x) \underset{n = 0}{\sum^{\infty}} \frac{{(\ln (x))}^{n}}{n! (n + 1)} + C$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum