I=∫(dx/((x^2 +2x+3)(√(x^2 +x+3)))) = ? my try..


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Question Number 111859 by ajfour last updated on 05/Sep/20

$I = \int \frac{d x}{(x^{2} + 2 x + 3) \sqrt{x^{2} + x + 3}} = ?$ $m y t r y . .$
Commented by Sarah85 last updated on 05/Sep/20

$my try :$ $(1 .) u = \ln (2 x + 1 + 2 \sqrt{x^{2} + x + 3})$ $x = \frac{e^{2 u} - {2 e}^{u} - 11}{{4 e}^{u}}$ $d x = \sqrt{x^{2} + x + 3} d u$ $16 \int \frac{e^{2 u}}{e^{4 u} + {4 e}^{3 u} + {14 e}^{2 u} - {44 e}^{u} + 121} d u$ $(2 .) v = e^{u}$ $u = \ln v$ $d u = \frac{d v}{e^{u}}$ $16 \int \frac{v}{v^{4} + 4 v^{3} + 14 v^{2} - 44 v + 121} d v$ $of course now we must decompose and {it}^{'} s$ $theoretically easy but hard to write out . . .$ $I {don}^{'} t think {there}^{'} s an alternative without$ $a 4^{th} - degree polynome . . .$
Commented by MJS_new last updated on 05/Sep/20

$good . I would add a 3^{rd} step$ $w = v + 1 \to d v = d w$ $16 \int \frac{w - 1}{w^{4} + 8 w^{2} - 64 w + 176} d w$ $the square factors are$ $w^{2} - 2 \sqrt{2 + 2 \sqrt{3}} w + 4 (2 + \sqrt{3} - 2 \sqrt{- 1 + \sqrt{3}})$ $and$ $w^{2} + 2 \sqrt{2 + 2 \sqrt{3}} w + 4 (2 + \sqrt{3} + 2 \sqrt{- 1 + \sqrt{3}})$ $\Rightarrow$ $16 \int \frac{w - 1}{w^{4} + 8 w^{2} - 64 w + 176} d w =$ $= \frac{\sqrt{3}}{6} \int \frac{\sqrt{- 1 + \sqrt{3}} w - 4 + 2 \sqrt{2 + 2 \sqrt{3}}}{w^{2} - 2 \sqrt{2 + 2 \sqrt{3}} w + 4 (2 + \sqrt{3} - 2 \sqrt{- 1 + \sqrt{3}})} d w -$ $- \frac{\sqrt{3}}{6} \int \frac{\sqrt{- 1 + \sqrt{3}} w + 4 + 2 \sqrt{2 + 2 \sqrt{3}}}{w^{2} + 2 \sqrt{2 + 2 \sqrt{3}} w + 4 (2 + \sqrt{3} + 2 \sqrt{- 1 + \sqrt{3}})} d w$ $and we can solve these using the formula for$ $\int \frac{a x + b}{x^{2} + c x + d} d x$ $if we must; i . e . if our survival depends on it$
Commented by mathdave last updated on 06/Sep/20

$u c a n o n l y s u r v i v e a w a y o u t o n l y b y$ $u s i n g f e r r a r i t r i c k o r i d e a t o$ $d e c o m p o s e d d e g r e e 4 t o d i f f e r e n c e o f a$ $t w o s q u a r e$
Commented by mathdave last updated on 06/Sep/20

$s h o u l d i h e l p u b r e a k t h i s 4 d e g r e e o f$ $t h i s e q u a t i o n$
Commented by mathdave last updated on 06/Sep/20

$w h o t o l d u d a t t h e r e i s n o a l t e r n a t i v e$ $w i t h o u t d e g r e e 4 p o l y n o m i a l, i f u$ $w a n n a k w n c h e c k m y w o r k i n g t o$ $s e e d a t$
Commented by MJS_new last updated on 06/Sep/20

$no help needed, thank you very much .$ $as you can see (if you take a closer look) I$ $already decomposed the 4^{th} degree . all {that}^{'} s$ $left is inserting the constants in the well$ $known formula . no need to re - invent the$ $wheel .$
Commented by mathdave last updated on 06/Sep/20

$a n y h o w o r w a t e v e r$
	Answered by mathdave last updated on 06/Sep/20

	$s o l u t i o n t o$ $l e t I = \int \frac{d x}{(x^{2} + 2 x + 3) \sqrt{x^{2} + x + 3}} p u t x = \frac{\sqrt{3} - \sqrt{3} t}{1 + t} a n d d x = - \frac{2 \sqrt{3}}{{(1 + t)}^{2}} d t$ $I = \int \frac{- 2 \sqrt{3} d t}{(\frac{3 {(1 - t)}^{2}}{{(1 + t)}^{2}} + 2 \sqrt{3} \frac{(1 - t)}{(1 + t)} + 3) \sqrt{\frac{3 {(1 - t)}^{2}}{{(1 + t)}^{2}} + \sqrt{3} (\frac{1 - t}{1 + t}) + 3}}$ $b y s i m p l i f y w e h a v e t h a t$ $I = \int \frac{- 2 \sqrt{3} (1 + t) d t}{(6 - 2 \sqrt{3}) t^{2} + (6 + 2 \sqrt{3})) \sqrt{(6 - \sqrt{3}) t^{2} + (6 + \sqrt{3})}}$ $l e t a = 6 - 2 \sqrt{3}, b = 6 + 2 \sqrt{3}, c = 6 - \sqrt{3}, d = 6 + \sqrt{3}$ $\int \frac{- 2 \sqrt{3} (1 + t) d t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} = - 2 \sqrt{3} \int \frac{d t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} - 2 \sqrt{3} \int \frac{t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} d t$ $L e t A = \int \frac{d t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} l e t y = \frac{t}{\sqrt{{c t}^{2} + d}}, t^{2} = \frac{y^{2} d}{1 - y^{2} c}, d t = \frac{\sqrt{d}}{{(1 - y^{2} c)}^{\frac{3}{2}}} d y$ $A = \int \frac{y \sqrt{d}}{\frac{{(1 - y^{2} c)}^{\frac{3}{2}}}{[a (\frac{y^{2} d}{1 - y^{2} c}) + b] \sqrt{\frac{y^{2} d}{1 - y^{2} c}}}} d y =$ $A = \int \frac{d y}{({a y}^{2} d + b - y^{2} b c) (1 - y^{2} c)}$ $A = \int \frac{(b c - a d) d y}{a d [(b c - a d) y^{2} - b]} - \int \frac{c d y}{a d ({c y}^{2} - 1)}$ $A = \frac{a d - b c}{a d} \int \frac{d y}{(a d - b c) y^{2} + b} + \frac{1}{a d} \int \frac{d y}{\frac{1}{c} - y^{2}}$ $A = \frac{a d - b c}{a d (a d - b c)} \int \frac{d y}{y^{2} + {[\sqrt{\frac{b}{a d - b c}}]}^{2}} + \frac{1}{a d} \int \frac{d y}{{(\frac{1}{\sqrt{c}})}^{2} - y^{2}}$ $A = \frac{1}{a d} \sqrt{\frac{a d - b c}{b}} \tan^{- 1} [\frac{y \sqrt{a d - b c}}{\sqrt{b}}] + \frac{1}{a d} \times \frac{\sqrt{c}}{2} \ln [\frac{1}{\frac{\sqrt{c}}{\frac{1}{\sqrt{c}} - y}} + y]$ $A = \frac{1}{a d} \sqrt{\frac{a d - b c}{b}} \tan^{- 1} [\frac{y \sqrt{a d - b c}}{\sqrt{b}}] + \frac{\sqrt{c}}{2 a d} \ln [\frac{1 + y \sqrt{c}}{1 - y \sqrt{c}}]$ $A = \frac{1}{a d} \sqrt{\frac{a d - b c}{b}} \tan^{- 1} [\frac{t \sqrt{a d - b c}}{\sqrt{{c b t}^{2} + b d}}] + \frac{\sqrt{c}}{2 a d} \ln [\frac{\sqrt{{c t}^{2} + d} + t \sqrt{c}}{\sqrt{{c t}^{2} + d} - t \sqrt{c}}] . . . . (1)$ $t h e n B = \int \frac{t d t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} = \frac{1}{2} \int \frac{d (t^{2})}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}}$ $B = \frac{1}{2} \int \frac{d u}{(a u + b) \sqrt{c u + d}} (p u t u = t^{2}, p u t p^{2} = c u + d, 2 p d p = c d u)$ $B = \frac{1}{2 c} \int \frac{2 p d p}{[a (\frac{- d + p^{2}}{c}) + b] p} = \frac{c}{c} \int \frac{d p}{({a p}^{2} - a d + b c)}$ $B = \int \frac{d p}{{a p}^{2} + (b c - a d)} = \frac{1}{a} \int \frac{d p}{p^{2} + (\frac{b c - a d}{a})}$ $B = \frac{1}{a} \int \frac{d p}{p^{2} + (\sqrt{\frac{b c - a d}{a}})} = \frac{\sqrt{a}}{a \sqrt{b c - a d}} \tan^{- 1} [\frac{p d p}{\sqrt{b c - a d}}]$ $B = \frac{1}{\sqrt{a} \sqrt{b c - a d}} \tan^{- 1} [\frac{\sqrt{a c u + d a}}{\sqrt{b c - a d}}] + k$ $B = \frac{1}{\sqrt{a b c - a^{2} d}} \tan^{- 1} [\frac{\sqrt{{a c t}^{2} + a d}}{\sqrt{b c - a d}}] + k . . . . . . . (2)$ $t h e n$ $I = - \frac{2 \sqrt{3}}{a d} \sqrt{\frac{a d - b c}{b}} \tan^{- 1} [\frac{t \sqrt{a d - b c}}{\sqrt{{c b t}^{2} + b d}}] - \frac{\sqrt{3} \sqrt{c}}{a d} \ln [\frac{\sqrt{{c t}^{2} + d} + t \sqrt{c}}{\sqrt{{c t}^{2} + d} - t \sqrt{c}}]$ $- \frac{2 \sqrt{3}}{\sqrt{a b c - a^{2} d}} \tan^{- 1} [\frac{\sqrt{{a c t}^{2} + a d}}{\sqrt{b c - a d}}] + k$ $b u t t = \frac{\sqrt{3} - x}{\sqrt{3} + x}$ $I = - \frac{2 \sqrt{3}}{a d} \sqrt{\frac{a d - b c}{b}} \tan^{- 1} [\frac{(\sqrt{3} - x) \sqrt{a d - b c}}{(\sqrt{3} + x) \sqrt{c b {(\frac{\sqrt{3} - x}{\sqrt{3} + x})}^{2} + b d}}]$ $- \frac{\sqrt{3} \sqrt{c}}{a d} \ln [\frac{\sqrt{c {(\frac{\sqrt{3} - x}{\sqrt{3} + x})}^{2} + d} + (\frac{\sqrt{3} - x}{\sqrt{3} + x}) \sqrt{c}}{\sqrt{c {(\frac{\sqrt{3} - x}{\sqrt{3} + x})}^{2} + d} - (\frac{\sqrt{3} - x}{\sqrt{3} + x}) \sqrt{c}}]$ $- \frac{2 \sqrt{3}}{\sqrt{a b c - a^{2} d}} \tan^{- 1} [\frac{\sqrt{a c {(\frac{\sqrt{3} - x}{\sqrt{3} + x})}^{2} + a d}}{\sqrt{b c - a d}}] + k$ $w h e r e a = 6 - 2 \sqrt{3}, b = 6 + 2 \sqrt{3}, c = 6 - \sqrt{3}, d = 6 + \sqrt{3}$ $b y m a t h d a v e (05 / 09 / 2020)$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
	Answered by ajfour last updated on 06/Sep/20
	$I=∫(dx/({(x+1)^2 +2}(√((x+(1/2))^2 +((11)/4))))) let x+(1/2)=((√(11))/2)tan θ I=∫((((√(11))/2)sec^2 θdθ)/({((((√(11))tan θ+1)^2 )/4)+2}((√(11))/2)sec θ)) =4∫((sec θdθ)/(((√(11))tan θ+1)^2 +8)) =4∫((cos θdθ)/(11sin^2 θ+(√(11))sin 2θ+9cos^2 θ)) =4∫((cos θdθ)/(11+(√(11))sin 2θ−1−cos 2θ)) I =4∫((cos θdθ)/(10−2(√3)cos (2θ+tan^(−1) (√(11))))) say θ+β=φ ; β=(1/2)tan^(−1) (√(11)) I=4∫((cos (φ−β)dφ)/(10−2(√3)cos (2φ))) I =4cos β∫((cos φdφ)/(10−2(√3)+4(√3)sin^2 φ)) −4sin β∫(((−sin φ)dφ)/(10+2(√3)−4(√3)cos^2 φ))+c −−−−−−−−−−−−−−−−−− I=((cos β)/( (√3)))×(√((4(√3))/(10−2(√3))))tan^(−1) [(√((4(√3))/(10−2(√3))))sin (θ+β)] −((sin β)/( (√3)))×(1/2)(√((4(√3))/(10+2(√3))))ln ∣(((√((10+2(√3))/(4(√3))))+cos (θ+β))/( (√((10+2(√3))/(4(√3))))−cos (θ+β)))∣+c θ=tan^(−1) (((2x+1)/( (√(11))))) ; β=(1/2)tan^(−1) (√(11)) ★−−−−−−−−−−−−−−−−★ rough work tan 2β=((2tan β)/(1−tan^2 β))=(√(11)) ⇒ (√(11))tan^2 β+2tan β−(√(11))=0 tan β=((2(√3)−1)/( (√(11)))) sin β=((2(√3)−1)/( (√(24−4(√3))))) = (√((2(√3)−1)/(4(√3)))) cos β= ((√(11))/( (√(4(√3)))(√(2(√3)−1)))) _____________________________ I=(((3(√2)+(√6))/(12)))tan^(−1) [(√((4(√3))/(10−2(√3))))sin φ] −((√(3(√3)−3))/(12)) ln ∣(((√(10+2(√3)))+(√(4(√3)))cos φ)/( (√(10+2(√3)))−(√(4(√3)))cos φ))∣+c ............................................. ∀ tan φ=(((2(√3)+1)/( (√(11)))))(((x+(√3))/( (√3)−x))) _____________________________ rough work tan (θ+β)=((((2x+1)/( (√(11))))+((2(√3)−1)/( (√(11)))))/(1−(((2x+1)(2(√3)−1))/(11)))) =(((√(11))(x+(√3)))/(6−(√3)−(2(√3)−1)x)) =(((√(11))(x+(√3)))/((2(√3)−1)((√3)−x))) −−−−−−−−−−−−−−$
	$I = \int \frac{d x}{{{(x + 1)}^{2} + 2} \sqrt{{(x + \frac{1}{2})}^{2} + \frac{11}{4}}}$ $l e t x + \frac{1}{2} = \frac{\sqrt{11}}{2} \tan θ$ $I = \int \frac{\frac{\sqrt{11}}{2} \sec^{2} θ d θ}{{\frac{{(\sqrt{11} \tan θ + 1)}^{2}}{4} + 2} \frac{\sqrt{11}}{2} \sec θ}$ $= 4 \int \frac{\sec θ d θ}{{(\sqrt{11} \tan θ + 1)}^{2} + 8}$ $= 4 \int \frac{\cos θ d θ}{11 \sin^{2} θ + \sqrt{11} \sin 2 θ + 9 \cos^{2} θ}$ $= 4 \int \frac{\cos θ d θ}{11 + \sqrt{11} \sin 2 θ - 1 - \cos 2 θ}$ $I = 4 \int \frac{\cos θ d θ}{10 - 2 \sqrt{3} \cos (2 θ + \tan^{- 1} \sqrt{11})}$ $s a y θ + β = ϕ; β = \frac{1}{2} \tan^{- 1} \sqrt{11}$ $I = 4 \int \frac{\cos (ϕ - β) d ϕ}{10 - 2 \sqrt{3} \cos (2 ϕ)}$ $I = 4 \cos β \int \frac{\cos ϕ d ϕ}{10 - 2 \sqrt{3} + 4 \sqrt{3} \sin^{2} ϕ}$ $- 4 \sin β \int \frac{(- \sin ϕ) d ϕ}{10 + 2 \sqrt{3} - 4 \sqrt{3} \cos^{2} ϕ} + c$ $- - - - - - - - - - - - - - - - - -$ $I = \frac{\cos β}{\sqrt{3}} \times \sqrt{\frac{4 \sqrt{3}}{10 - 2 \sqrt{3}}} \tan^{- 1} [\sqrt{\frac{4 \sqrt{3}}{10 - 2 \sqrt{3}}} \sin (θ + β)]$ $- \frac{\sin β}{\sqrt{3}} \times \frac{1}{2} \sqrt{\frac{4 \sqrt{3}}{10 + 2 \sqrt{3}}} \ln ∣ \frac{\sqrt{\frac{10 + 2 \sqrt{3}}{4 \sqrt{3}}} + \cos (θ + β)}{\sqrt{\frac{10 + 2 \sqrt{3}}{4 \sqrt{3}}} - \cos (θ + β)} ∣ + c$ $θ = \tan^{- 1} (\frac{2 x + 1}{\sqrt{11}}); β = \frac{1}{2} \tan^{- 1} \sqrt{11}$ $★ - - - - - - - - - - - - - - - - ★$ $r o u g h w o r k$ $\tan 2 β = \frac{2 \tan β}{1 - \tan^{2} β} = \sqrt{11}$ $\Rightarrow \sqrt{11} \tan^{2} β + 2 \tan β - \sqrt{11} = 0$ $\tan β = \frac{2 \sqrt{3} - 1}{\sqrt{11}}$ $\sin β = \frac{2 \sqrt{3} - 1}{\sqrt{24 - 4 \sqrt{3}}} = \sqrt{\frac{2 \sqrt{3} - 1}{4 \sqrt{3}}}$ $\cos β = \frac{\sqrt{11}}{\sqrt{4 \sqrt{3}} \sqrt{2 \sqrt{3} - 1}}$ $_____________________________$ $I = (\frac{3 \sqrt{2} + \sqrt{6}}{12}) \tan^{- 1} [\sqrt{\frac{4 \sqrt{3}}{10 - 2 \sqrt{3}}} \sin ϕ]$ $- \frac{\sqrt{3 \sqrt{3} - 3}}{12} \ln ∣ \frac{\sqrt{10 + 2 \sqrt{3}} + \sqrt{4 \sqrt{3}} \cos ϕ}{\sqrt{10 + 2 \sqrt{3}} - \sqrt{4 \sqrt{3}} \cos ϕ} ∣ + c$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $\forall \tan ϕ = (\frac{2 \sqrt{3} + 1}{\sqrt{11}}) (\frac{x + \sqrt{3}}{\sqrt{3} - x})$ $_____________________________$ $r o u g h w o r k$ $\tan (θ + β) = \frac{\frac{2 x + 1}{\sqrt{11}} + \frac{2 \sqrt{3} - 1}{\sqrt{11}}}{1 - \frac{(2 x + 1) (2 \sqrt{3} - 1)}{11}}$ $= \frac{\sqrt{11} (x + \sqrt{3})}{6 - \sqrt{3} - (2 \sqrt{3} - 1) x} = \frac{\sqrt{11} (x + \sqrt{3})}{(2 \sqrt{3} - 1) (\sqrt{3} - x)}$ $- - - - - - - - - - - - - -$
	Commented by ajfour last updated on 06/Sep/20

	${I s n}^{'} t t h i s a l r i g h t ?$
	Commented by MJS_new last updated on 06/Sep/20

	$I^{'} ll check it tomorrow$
	Commented by mathdave last updated on 06/Sep/20

	$t h i s c a n t b t r u e$
	Commented by MJS_new last updated on 06/Sep/20

	${it}^{'} s right, Sir Aifour . can you finish it ?$
	Commented by ajfour last updated on 06/Sep/20

	$t h a n k y o u S i r!$
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