find ∫ ((ln(x+(√(x^2 −1))))/(√(x^2 −1))) dx 2) calculate ∫_2 ^5 ((ln(x+(√(x^2 −1)))/(√(x^2 −1)))dx


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Question Number 39833 by math khazana by abdo last updated on 12/Jul/18

$f i n d \int \frac{l n (x + \sqrt{x^{2} - 1})}{\sqrt{x^{2} - 1}} d x$ $2) c a l c u l a t e \int_{2}^{5} \frac{l n (x + \sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1}} d x$
Commented by abdo mathsup 649 cc last updated on 12/Jul/18
$let I = ∫ ((ln(x+(√(x^2 −1))))/(√(x^2 −1)))dx let integrate by parts u^′ =(1/(√(x^2 −1))) and v(x)= ln(x+(√(x^2 −1))) I =ln^2 (x+(√(x^2 −1))) −∫ ((ln(x+(√(x^2 −1))))/(√(x^2 −1)))dx =ln^2 (x+(√(x^2 −1)))−I ⇒ 2I =ln^2 (x+(√(x^2 −1))) ⇒ I =(1/2)ln^2 (x+(√(x^2 −1))) 2) ∫_2 ^5 ((ln(x+(√(x^2 −1))))/(√(x^2 −1)))dx=[(1/2)ln^2 (x+(√(x^2 −1)))]_2 ^5 =(1/2){ln^2 (5 +(√(24)))−ln^2 (2+(√3))}$
$l e t I = \int \frac{l n (x + \sqrt{x^{2} - 1})}{\sqrt{x^{2} - 1}} d x l e t i n t e g r a t e b y p a r t s$ $u^{^{'}} = \frac{1}{\sqrt{x^{2} - 1}} a n d v (x) = l n (x + \sqrt{x^{2} - 1})$ $I = {l n}^{2} (x + \sqrt{x^{2} - 1}) - \int \frac{l n (x + \sqrt{x^{2} - 1})}{\sqrt{x^{2} - 1}} d x$ $= {l n}^{2} (x + \sqrt{x^{2} - 1}) - I \Rightarrow$ $2 I = {l n}^{2} (x + \sqrt{x^{2} - 1}) \Rightarrow I = \frac{1}{2} {l n}^{2} (x + \sqrt{x^{2} - 1})$ $2) \int_{2}^{5} \frac{l n (x + \sqrt{x^{2} - 1})}{\sqrt{x^{2} - 1}} d x = {[\frac{1}{2} {l n}^{2} (x + \sqrt{x^{2} - 1})]}_{2}^{5}$ $= \frac{1}{2} {{l n}^{2} (5 + \sqrt{24}) - {l n}^{2} (2 + \sqrt{3})}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18

	$t = x + \sqrt{x^{2} - 1}$ $\frac{d t}{d x} = 1 + \frac{2 x}{2 \sqrt{x^{2} - 1}} = \frac{x + \sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1}}$ $\int l n t d t$ $t l n t - t + c$ $(x + \sqrt{x^{2} - 1}) l n (x + \sqrt{x^{2} - 1}) - (x + \sqrt{x^{2} - 1}) + c$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18

	$t = x + \sqrt{x^{2} - 1}$ $\frac{d t}{d x} = 1 + \frac{2 x}{2 \sqrt{x^{2} - 1}} = \frac{x + \sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1}}$ $\int l n t d t$ $t l n t - t + c$ $(x + \sqrt{x^{2} - 1}) l n (x + \sqrt{x^{2} - 1}) - (x + \sqrt{x^{2} - 1}) + c$
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