find ∫ (dx/((x+(√(x−1)))^2 ))


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Question Number 89317 by abdomathmax last updated on 16/Apr/20

$f i n d \int \frac{d x}{{(x + \sqrt{x - 1})}^{2}}$
Commented by mathmax by abdo last updated on 17/Apr/20
$parametric method let f(a) =∫ (dx/(a+x+(√(x−1)))) we have f^′ (a) =−∫ (dx/((a+x+(√(x−1)))^2 )) ⇒∫(dx/((a+x+(√(x−1)))^2 ))=−f^′ (a) and ∫ (dx/((x+(√(x−1)))^2 )) =−f^′ (0) changement (√(x−1))=t give x−1=t^2 ⇒ f(a) =∫ ((2tdt)/(a+1+t^2 +t)) =2 ∫ ((tdt)/(t^2 +t +a+1)) =∫((2t+1−1)/(t^2 +t+a+1))dt =ln∣t^2 +t +a+1∣−∫ (dt/(t^2 +t+1+a)) we have ∫ (dt/(t^2 +t+1 +a)) =∫ (dt/((t+(1/2))^2 +a+(3/4))) =_(t+(1/2)=(√(a+(3/4)))u) =(1/((a+(3/4)))) ∫ (1/(1+u^2 ))×(√(a+(3/4)))du =(1/(√(a+(3/4)))) arctan(((2t+1)/(√(4a+3)))) +C =(2/(√(4a+3))) arctan(((2(√(x−1))+1)/(√(4a+3)))) +C ⇒ f^′ (a) =2{(4a+3)^(−(1/2)) arctan(((2(√(x−1))+1)/(√(4a+3))))}^((1)) =−4(4a+3)^(−(3/2)) arctan(((2(√(x−1))+1)/(√(4a+3))))+2 (4a+3)^(−(1/2)) ×(((((2(√(x−1))+1)/(√(4a+3))))^′ )/(1+(((2(√(x−1))+1)/(√(4a+3))))^2 )) =−4(4a+3)^(−(3/2)) arctan(((2(√(x−1))+1)/(√(4a+3))))+2(4a+3)^(−(1/2)) ×(1/((√(4a+3))(√(x−1))(1+(((2(√(x−1))+1)^2 )/(4a+3))))) f^′ (0) =−4(3)^(−(3/2)) arctan(((2(√(x−1))+1)/(√3))) +2(3)^(−(1/2)) ×(1/((√3)(√(x−1))(1+(((2(√(x−1))+1)^2 )/3)))) I =−f^′ (0)$
$p a r a m e t r i c m e t h o d l e t f (a) = \int \frac{d x}{a + x + \sqrt{x - 1}}$ $w e h a v e f^{^{'}} (a) = - \int \frac{d x}{{(a + x + \sqrt{x - 1})}^{2}} \Rightarrow \int \frac{d x}{{(a + x + \sqrt{x - 1})}^{2}} = - f^{^{'}} (a)$ $a n d \int \frac{d x}{{(x + \sqrt{x - 1})}^{2}} = - f^{^{'}} (0)$ $c h a n g e m e n t \sqrt{x - 1} = t g i v e x - 1 = t^{2} \Rightarrow$ $f (a) = \int \frac{2 t d t}{a + 1 + t^{2} + t} = 2 \int \frac{t d t}{t^{2} + t + a + 1}$ $= \int \frac{2 t + 1 - 1}{t^{2} + t + a + 1} d t = l n ∣ t^{2} + t + a + 1 ∣ - \int \frac{d t}{t^{2} + t + 1 + a} w e h a v e$ $\int \frac{d t}{t^{2} + t + 1 + a} = \int \frac{d t}{{(t + \frac{1}{2})}^{2} + a + \frac{3}{4}} =_{t + \frac{1}{2} = \sqrt{a + \frac{3}{4}} u}$ $= \frac{1}{(a + \frac{3}{4})} \int \frac{1}{1 + u^{2}} \times \sqrt{a + \frac{3}{4}} d u$ $= \frac{1}{\sqrt{a + \frac{3}{4}}} a r c t a n (\frac{2 t + 1}{\sqrt{4 a + 3}}) + C$ $= \frac{2}{\sqrt{4 a + 3}} a r c t a n (\frac{2 \sqrt{x - 1} + 1}{\sqrt{4 a + 3}}) + C \Rightarrow$ $f^{^{'}} (a) = 2 {{(4 a + 3)}^{- \frac{1}{2}} a r c t a n (\frac{2 \sqrt{x - 1} + 1}{\sqrt{4 a + 3}})}^{(1)}$ $= - 4 {(4 a + 3)}^{- \frac{3}{2}} a r c t a n (\frac{2 \sqrt{x - 1} + 1}{\sqrt{4 a + 3}}) + 2 {(4 a + 3)}^{- \frac{1}{2}} \times \frac{{(\frac{2 \sqrt{x - 1} + 1}{\sqrt{4 a + 3}})}^{^{'}}}{1 + {(\frac{2 \sqrt{x - 1} + 1}{\sqrt{4 a + 3}})}^{2}}$ $= - 4 {(4 a + 3)}^{- \frac{3}{2}} a r c t a n (\frac{2 \sqrt{x - 1} + 1}{\sqrt{4 a + 3}}) + 2 {(4 a + 3)}^{- \frac{1}{2}} \times \frac{1}{\sqrt{4 a + 3} \sqrt{x - 1} (1 + \frac{{(2 \sqrt{x - 1} + 1)}^{2}}{4 a + 3})}$ $f^{^{'}} (0) = - 4 {(3)}^{- \frac{3}{2}} a r c t a n (\frac{2 \sqrt{x - 1} + 1}{\sqrt{3}}) + 2 {(3)}^{- \frac{1}{2}} \times \frac{1}{\sqrt{3} \sqrt{x - 1} (1 + \frac{{(2 \sqrt{x - 1} + 1)}^{2}}{3})}$ $I = - f^{^{'}} (0)$
Commented by mathmax by abdo last updated on 17/Apr/20

$f o r g i v e f^{^{'}} (a) = \frac{1}{t^{2} + t + a + 1} + 4 {(4 a + 3)}^{- \frac{3}{2}} a r c t a n (\frac{2 \sqrt{x - 1} + 1}{\sqrt{4 a + 3}})$ $- 2 {(4 a + 3)}^{- \frac{1}{2}} \times \frac{1}{\sqrt{4 a + 3} \sqrt{x - 1} (1 + \frac{{(2 \sqrt{x - 1} + 1)}^{2}}{4 a + 3})} \Rightarrow$ $f^{^{'}} (0) = \frac{1}{t^{2} + t + 1} + 4 {(3)}^{- \frac{3}{2}} a r c t a n (\frac{2 \sqrt{x - 1} + 1}{\sqrt{3}})$ $- 2 {(3)}^{- \frac{1}{2}} \times \frac{1}{\sqrt{3} \sqrt{x - 1} (1 + \frac{{(2 \sqrt{x - 1} + 1)}^{2}}{3})}$
	Answered by MJS last updated on 16/Apr/20

	$\int \frac{d x}{{(x + \sqrt{x - 1})}^{2}} =$ $[t = \sqrt{x - 1} \to d x = 2 \sqrt{x - 1} d t]$ $= 2 \int \frac{t}{{(t^{2} + t + 1)}^{2}} d t =$ $[Ostrogradski]$ $= - \frac{2 (t + 2)}{3 (t^{2} + t + 1)} - \frac{2}{3} \int \frac{d t}{t^{2} + t + 1} =$ $= - \frac{2 (t + 2)}{3 (t^{2} + t + 1)} - \frac{4 \sqrt{3}}{9} \arctan \frac{2 t + 1}{\sqrt{3}} =$ $= - \frac{2 (2 + \sqrt{x - 1})}{3 (x + \sqrt{x - 1})} - \frac{4 \sqrt{3}}{9} \arctan \frac{1 + 2 \sqrt{x - 1}}{\sqrt{3}} + C$
	Commented by mathmax by abdo last updated on 17/Apr/20

	$t h a n k y o u s i r .$
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