∫_2 ^( e) ((1/(ln x))−(1/(ln^2 x))) dx?


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Question Number 87669 by john santu last updated on 05/Apr/20

$\int_{2}^{e} (\frac{1}{\ln x} - \frac{1}{\ln^{2} x}) dx ?$
Commented by mathmax by abdo last updated on 05/Apr/20
$I =∫_2 ^e (((lnx−1)/(ln^2 x)))dx vhangement lnx =t give I =∫_(ln2) ^1 ((t−1)/t^2 )e^t dt =∫_(ln2) ^1 (((t−1)e^t )/t^2 )dt =_(by parts) [−(1/t)(t−1)e^t ]_(ln(2)) ^1 +∫_(ln(2)) ^1 (1/t){ e^t +(t−1)e^t }dt =(((ln2−1)2)/(ln2)) +∫_(ln2) ^1 e^t dt =2−(2/(ln2)) +e−2 =e−(2/(ln2))$
$I = \int_{2}^{e} (\frac{l n x - 1}{{l n}^{2} x}) d x v h a n g e m e n t l n x = t g i v e$ $I = \int_{l n 2}^{1} \frac{t - 1}{t^{2}} e^{t} d t = \int_{l n 2}^{1} \frac{(t - 1) e^{t}}{t^{2}} d t =_{b y p a r t s} {[- \frac{1}{t} (t - 1) e^{t}]}_{l n (2)}^{1}$ $+ \int_{l n (2)}^{1} \frac{1}{t} {e^{t} + (t - 1) e^{t}} d t = \frac{(l n 2 - 1) 2}{l n 2} + \int_{l n 2}^{1} e^{t} d t$ $= 2 - \frac{2}{l n 2} + e - 2 = e - \frac{2}{l n 2}$
Commented by Ar Brandon last updated on 05/Apr/20

$I t s e e m s y o u m a d e a n e r r o r w h i l e i n t e g r a t i n g b y p a r t .$ ${I t}^{'} s n o t - \frac{1}{t} (t - 1) e^{t} b u t (l n t + \frac{1}{t}) e^{t}$ $M a y b e y o u s h o u l d v e r i f y y o u r s o l u t i o n .$
Commented by abdomathmax last updated on 05/Apr/20

$n o s i r b y p s r t s i h a v e t a k e n u^{^{'}} = \frac{1}{t^{2}}$ $a n d v = (t - 1) e^{t}$
Commented by Ar Brandon last updated on 05/Apr/20

$O K b r o . {I t}^{'} s c l e a r, t h a n k s .$
	Answered by TANMAY PANACEA. last updated on 05/Apr/20

	$\int_{2}^{e} \frac{l n x - 1}{{(l n x)}^{2}} d x$ $\int_{2}^{e} \frac{l n x \times \frac{d x}{d x} - x \times \frac{d}{d x} (l n x)}{{(l n x)}^{2}} d x$ $\int_{2}^{e} \frac{d}{d x} (\frac{x}{l n x}) d x$ $\int_{2}^{e} d (\frac{x}{l n x}) =∣ \frac{x}{l n x} ∣_{2}^{e} = \frac{e}{l n e} - \frac{2}{l n 2} = e - \frac{2}{l n 2} (c o r r e c t e d)$
	Commented by TANMAY PANACEA. last updated on 05/Apr/20

	$t h a k y o u s i r$
	Commented by john santu last updated on 05/Apr/20

	$waw === your method$ $funtastic sir$
	Commented by Ar Brandon last updated on 05/Apr/20

	$W o w!$
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