find ∫ ((x+3)/(√(x^2 +x −1)))dx


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Question Number 35428 by abdo.msup.com last updated on 18/May/18

$f i n d \int \frac{x + 3}{\sqrt{x^{2} + x - 1}} d x$
Commented by prof Abdo imad last updated on 19/May/18
$let put I = ∫ ((x+3)/(√(x^2 +x−1)))dx I = (1/2) ∫ ((2x+6)/(√(x^2 +x−1)))dx =(1/2)∫ ((2x+1)/(√(x^2 +x−1)))dx +(5/2) ∫ (dx/(√(x^2 +x−1))) but ∫ ((2x+1)/(2(√(x^2 +x−1))))dx =(√(x^2 +x−1)) +λ x^2 +x−1 = x^2 +2x(1/2) +(1/4) −1−(1/4) =(x+(1/2))^2 −(5/4) and cjsngement x+(1/2) =((√5)/2)cht give ∫ (dx/(√(x^2 +x−1))) = ((√5)/2)∫ ((sht dt)/(√((5/4)(ch^2 t−1)))) = ((√5)/2) .(2/(√5)) ∫ ((sht)/(sht))dt = t +c but 2x+1=(√5) cht⇒ cht =((2x+1)/(√5)) ⇒ t =argch(((2x+1)/(√5))) =ln( ((2x+1)/(√5)) +(√({((2x+1)/(√5))}^2 −1)) ) ⇒ I = (√(x^2 +x−1)) +(5/2)ln{ ((2x+1)/(√5)) +(√((((2x+1)/(√5)))^2 −1))) +λ$
$l e t p u t I = \int \frac{x + 3}{\sqrt{x^{2} + x - 1}} d x$ $I = \frac{1}{2} \int \frac{2 x + 6}{\sqrt{x^{2} + x - 1}} d x$ $= \frac{1}{2} \int \frac{2 x + 1}{\sqrt{x^{2} + x - 1}} d x + \frac{5}{2} \int \frac{d x}{\sqrt{x^{2} + x - 1}} b u t$ $\int \frac{2 x + 1}{2 \sqrt{x^{2} + x - 1}} d x = \sqrt{x^{2} + x - 1} + λ$ $x^{2} + x - 1 = x^{2} + 2 x \frac{1}{2} + \frac{1}{4} - 1 - \frac{1}{4}$ $= {(x + \frac{1}{2})}^{2} - \frac{5}{4} a n d c j s n g e m e n t x + \frac{1}{2} = \frac{\sqrt{5}}{2} c h t$ $g i v e \int \frac{d x}{\sqrt{x^{2} + x - 1}} = \frac{\sqrt{5}}{2} \int \frac{s h t d t}{\sqrt{\frac{5}{4} ({c h}^{2} t - 1)}}$ $= \frac{\sqrt{5}}{2} . \frac{2}{\sqrt{5}} \int \frac{s h t}{s h t} d t = t + c b u t 2 x + 1 = \sqrt{5} c h t \Rightarrow$ $c h t = \frac{2 x + 1}{\sqrt{5}} \Rightarrow t = a r g c h (\frac{2 x + 1}{\sqrt{5}})$ $= l n (\frac{2 x + 1}{\sqrt{5}} + \sqrt{{\frac{2 x + 1}{\sqrt{5}}}^{2} - 1}) \Rightarrow$ $I = \sqrt{x^{2} + x - 1} + \frac{5}{2} l n {\frac{2 x + 1}{\sqrt{5}} + \sqrt{{(\frac{2 x + 1}{\sqrt{5}})}^{2} - 1}) + λ$
	Answered by ajfour last updated on 19/May/18

	$I = \frac{1}{2} \int \frac{2 x + 1}{\sqrt{x^{2} + x - 1}} d x + \frac{5}{2} \int \frac{d x}{\sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{\sqrt{5}}{2})}^{2}}}$ $= \sqrt{x^{2} + x - 1} + \frac{5}{2} \ln ∣ x + \frac{1}{2} + \sqrt{x^{2} + x - 1} ∣ + c .$
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