solve ∫(dx/((x^2 +2x+3)(√(x^2 +x+3))))


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Question Number 111615 by mathdave last updated on 04/Sep/20

$s o l v e$ $\int \frac{d x}{(x^{2} + 2 x + 3) \sqrt{x^{2} + x + 3}}$
	Answered by mathmax by abdo last updated on 04/Sep/20
	$I =∫ (dx/((x^2 +2x+3)(√(x^2 +x+3)))) ⇒I =∫ (dx/((x^2 +x+3 +x)(√(x^2 +x+3)))) we have x^2 +x +3 =x^2 +((2x)/2) +(1/4) +3−(1/4) =(x+(1/2))^2 +((11)/4) we do the changement x+(1/2) =((√(11))/2) sh(t) ⇒ I =∫ (1/({((11)/4)sh^2 t +((11)/4) +((√(11))/2)sht−(1/2)}((√(11))/2)cht))((√(11))/2) sht dt =4∫ (dt/(11sh^2 t +11 +2(√(11))sht −2)) =4 ∫ (dt/(11sh^2 t +2(√(11))sht +9)) =4∫ (dt/(11((ch(2t)−1)/2)+2 (√(11))sh(t) +9)) 8 ∫ (dt/(11ch(2t)+4(√(11))sht +7)) =8 ∫ (dt/(11((e^(2t) +e^(−2t) )/2)+4(√(11))((e^t −e^(−t) )/2) +7)) =16 ∫ (dt/(11e^(2t) +11e^(−2t) +4(√(11))e^t −4(√(11))e^(−t) +14)) =_(e^t =z) 16 ∫ (dz/(z{ 11z^2 +11z^(−2) +4(√(11))z−4(√(11))z^(−1) +14})) =16 ∫ ((zdz)/(11z^4 +11 +4(√(11))z^3 −4(√(11))z+ 14z^2 )) =16 ∫ ((zdz)/(11z^4 +14z^2 +4(√(11))z^3 +11)) rest decomposition of f(z) =(z/(11z^4 +14z^2 +4(√(11))z^3 +11)) ....be continued....$
	$I = \int \frac{dx}{(x^{2} + 2 x + 3) \sqrt{x^{2} + x + 3}} \Rightarrow I = \int \frac{dx}{(x^{2} + x + 3 + x) \sqrt{x^{2} + x + 3}}$ $we have x^{2} + x + 3 = x^{2} + \frac{2 x}{2} + \frac{1}{4} + 3 - \frac{1}{4} = {(x + \frac{1}{2})}^{2} + \frac{11}{4}$ $we do the changement x + \frac{1}{2} = \frac{\sqrt{11}}{2} sh (t) \Rightarrow$ $I = \int \frac{1}{{\frac{11}{4} {sh}^{2} t + \frac{11}{4} + \frac{\sqrt{11}}{2} sht - \frac{1}{2}} \frac{\sqrt{11}}{2} cht} \frac{\sqrt{11}}{2} sht dt$ $= 4 \int \frac{dt}{{11 sh}^{2} t + 11 + 2 \sqrt{11} sht - 2} = 4 \int \frac{dt}{{11 sh}^{2} t + 2 \sqrt{11} sht + 9}$ $= 4 \int \frac{dt}{11 \frac{ch (2 t) - 1}{2} + 2 \sqrt{11} sh (t) + 9}$ $8 \int \frac{dt}{11 ch (2 t) + 4 \sqrt{11} sht + 7} = 8 \int \frac{dt}{11 \frac{e^{2 t} + e^{- 2 t}}{2} + 4 \sqrt{11} \frac{e^{t} - e^{- t}}{2} + 7}$ $= 16 \int \frac{dt}{{11 e}^{2 t} + {11 e}^{- 2 t} + 4 \sqrt{11} e^{t} - 4 \sqrt{11} e^{- t} + 14}$ $=_{e^{t} = z} 16 \int \frac{dz}{z {{11 z}^{2} + {11 z}^{- 2} + 4 \sqrt{11} z - 4 \sqrt{11} z^{- 1} + 14}}$ $= 16 \int \frac{zdz}{{11 z}^{4} + 11 + 4 \sqrt{11} z^{3} - 4 \sqrt{11} z + {14 z}^{2}}$ $= 16 \int \frac{zdz}{{11 z}^{4} + {14 z}^{2} + 4 \sqrt{11} z^{3} + 11} rest decomposition of$ $f (z) = \frac{z}{{11 z}^{4} + {14 z}^{2} + 4 \sqrt{11} z^{3} + 11} . . . . be continued . . . .$
	Answered by Her_Majesty last updated on 05/Sep/20

	$\int \frac{d x}{(x^{2} + 2 x + 3) \sqrt{x^{2} + x + 3}} =$ $u s e t = \frac{\sqrt{11}}{11} (2 x + 1 + 2 \sqrt{x^{2} + x + 3})$ $= \frac{16}{11} \int \frac{t}{t^{4} + \frac{4}{\sqrt{11}} t^{3} + \frac{14}{11} t^{2} - \frac{4}{\sqrt{11}} t + 1} d t$ $t^{4} + \frac{4}{\sqrt{11}} t^{3} + \frac{14}{11} t^{2} - \frac{4}{\sqrt{11}} t + 1 =$ $= (t^{2} + \frac{2}{\sqrt{11}} (1 + \sqrt{2 + 3 \sqrt{3}}) t + \frac{9 + 4 \sqrt{3} + 8 \sqrt{- 1 + \sqrt{3}} + 2 \sqrt{2 + 2 \sqrt{3}}}{11}) \times$ $\times (t^{2} + \frac{2}{\sqrt{11}} (1 - \sqrt{2 + 2 \sqrt{3}}) t + \frac{9 + 4 \sqrt{3} - 8 \sqrt{- 1 + \sqrt{3}} - 2 \sqrt{2 + 2 \sqrt{3}}}{11})$ $a n d I r e f u s e t o d e c o m p o s e i t, s o r r y$
	Commented by mathdave last updated on 05/Sep/20

	$u c y o u r l i f e w h y c a n t u c o n t i n u e t h e$ $s o l v e h a s u k n o w b o o k \cdot s o r r y u n o$ $n o t h i n g$
	Commented by mathdave last updated on 05/Sep/20

	$h a s u a n t a g o n i z e d m y w o r k i n g i$ $t h o u g h t y o u w i l l g i v e c o r r e c t e v a l u a t i o n$ $n o o n e i s m o n o p o l y o f k n o w l e d g e o o o$ $b u t a m n t a f r a i d t o s a y u c a n t$ $w i t h s t a n d m e i n m a t h e m a t i c s$
	Commented by Her_Majesty last updated on 05/Sep/20

	$S i r I b e g y o u t o a c c e p t t h a t$ $\frac{1}{a + b} \neq \frac{1}{a} + \frac{1}{b}$ ${t h a t}^{'} s a l l .$
	Commented by Her_Majesty last updated on 05/Sep/20
	$btw I can continue but I don′t want to. obviously I′d have to decompose the fraction and solve 2 integrals of the form ∫((at+b)/(t^2 +αt+β))dt but I don′t waste my time with these given constants; I won′t gain any new experience doing so.$
	$b t w I c a n c o n t i n u e b u t I {d o n}^{'} t w a n t t o .$ $o b v i o u s l y I^{'} d h a v e t o d e c o m p o s e t h e f r a c t i o n$ $a n d s o l v e 2 i n t e g r a l s o f t h e f o r m$ $\int \frac{a t + b}{t^{2} + α t + β} d t b u t I {d o n}^{'} t w a s t e m y t i m e w i t h$ $t h e s e g i v e n c o n s t a n t s; I {w o n}^{'} t g a i n a n y n e w$ $e x p e r i e n c e d o i n g s o .$
	Commented by mathdave last updated on 05/Sep/20

	$t e l l m e w h a t i d o n t k n o w f o r t h i s a m$ $101 % {(v e r y)}^{\infty} a n d s u p e r c o r r e c t$
	Commented by mathdave last updated on 05/Sep/20

	$u m o s t b v e r y v e r y s t u p i d w i t h t h a t b i g$ $m o u t h u j u s t s a i d u r e m a n n e r l e s s$
	Commented by Her_Majesty last updated on 05/Sep/20

	$i f y o u s o l v e a n i n t e g r a l t h e t e s t i s t h e$ $d e r i v a t i o n .$ $I h a v e s h o w n y o u r r e s u l t i s n o t c o r r e c t .$ $\int f (x) d x = F (x) + C \Leftrightarrow \frac{d}{d x} [F (x) + C] = f (x)$ $\frac{d}{d x} [F (x) + C] \neq f (x) \Rightarrow F (x) + C \neq \int f (x) d x$ $p r o v e t h a t I^{'} m w r o n g o r s h u t u p$
	Commented by mathdave last updated on 05/Sep/20

	$t h e n p r o c e e d t o d o t h e c o r r e c t t h i n g s l e t$ $m e a p p l a u d y o u f o r t h e g o o d w o r k . i f u$ $c a n t k e e p t h e f u c k u p l e t m e s e e v i e w s$
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