calculate L=∫_e^2 ^( e^3 ) ((ln(x)−1)/(xlnx)) dx Please detail if possible^


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 133942 by mathocean1 last updated on 25/Feb/21

$c a l c u l a t e L = \int_{e^{2}}^{e^{3}} \frac{l n (x) - 1}{x l n x} d x$ $P l e a s e d e t a i l i f p o s s i b l \overset{}{e}$
	Answered by Ñï= last updated on 25/Feb/21

	$L = \int_{e^{2}}^{e^{3}} \frac{l n x - 1}{x l n x} d x$ $= \int_{e^{2}}^{e^{3}} (\frac{1}{x} - \frac{1}{x l n x}) d x$ $= (l n x - l n l n x) ∣_{e^{2}}^{e^{3}}$ $= (3 - 2) - (l n 3 - l n 2)$ $= 1 - l n \frac{3}{2}$
	Answered by mnjuly1970 last updated on 25/Feb/21

	$l n (x) = u \Rightarrow \frac{1}{x} d x = d u$ $ϕ = \int_{2}^{3} \frac{u - 1}{u} d u$ ${[u - l n (u)]}_{2}^{3} = 3 - l n (3) - 2 + l n (2$ $= 1 - l n (\frac{3}{2}) = l n (\frac{2 e}{3}) . . ✓$
	Answered by Olaf last updated on 25/Feb/21

	$I = \int_{e^{2}}^{e^{3}} \frac{\ln x - 1}{x \ln x} d x$ $I = \int_{e^{2}}^{e^{3}} (\frac{\ln x + 1}{x \ln x} - 2 \frac{\frac{1}{x}}{\ln x}) d x$ $I = \int_{e^{2}}^{e^{3}} (\frac{d (x \ln x)}{x \ln x} - 2 \frac{d (\ln x)}{\ln x})$ $I = {[\ln (x \ln x) - 2 \ln (\ln x)]}_{e^{2}}^{e^{3}}$ $I = {[\ln (\frac{x}{\ln x})]}_{e^{2}}^{e^{3}}$ $I = \ln (\frac{e^{3}}{3}) - \ln (\frac{e^{2}}{2}) = \ln (\frac{2 e}{3}) = 1 + \ln \frac{2}{3}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum