∫(1/((x^2 +2x+5)^2 ))dx


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Question Number 44264 by pramid last updated on 25/Sep/18

$\int \frac{1}{{(x^{2} + 2 x + 5)}^{2}} dx$
Commented by maxmathsup by imad last updated on 25/Sep/18
$let I = ∫ (dx/((x^2 +2x +5)^2 )) ⇒ I =∫ (dx/({(x+1)^2 +4}^2 )) changement x+1 =2tanθ give I = ∫ ((2(1+tan^2 θ)dθ)/(4^2 {1+tan^2 θ }^2 )) =(1/8) ∫ (dθ/(1+tan^2 θ)) =(1/8) ∫cos^2 θ dθ = (1/(16)) ∫(1+cos(2θ))dθ+c =(θ/(16)) +(1/(32))sin(2θ)+c =((arctan(((x+1)/2)))/(16)) +(1/(32)) ((2tanθ)/(1+tan^2 θ))+c =(1/(16)) arctan(((x+1)/2)) +(1/8) ((x+1)/(2(1+(((x+1)/2))^2 ))) +c =(1/(16)) arctan(((x+1)/2)) +(1/(16)) ((x+1)/({1 +(((x+1)^2 )/4)})) +c .$
$l e t I = \int \frac{d x}{{(x^{2} + 2 x + 5)}^{2}} \Rightarrow I = \int \frac{d x}{{{(x + 1)}^{2} + 4}^{2}} c h a n g e m e n t x + 1 = 2 t a n θ g i v e$ $I = \int \frac{2 (1 + {t a n}^{2} θ) d θ}{4^{2} {1 + {t a n}^{2} θ}^{2}} = \frac{1}{8} \int \frac{d θ}{1 + {t a n}^{2} θ} = \frac{1}{8} \int {c o s}^{2} θ d θ$ $= \frac{1}{16} \int (1 + c o s (2 θ)) d θ + c = \frac{θ}{16} + \frac{1}{32} s i n (2 θ) + c = \frac{a r c t a n (\frac{x + 1}{2})}{16} + \frac{1}{32} \frac{2 t a n θ}{1 + {t a n}^{2} θ} + c$ $= \frac{1}{16} a r c t a n (\frac{x + 1}{2}) + \frac{1}{8} \frac{x + 1}{2 (1 + {(\frac{x + 1}{2})}^{2})} + c$ $= \frac{1}{16} a r c t a n (\frac{x + 1}{2}) + \frac{1}{16} \frac{x + 1}{{1 + \frac{{(x + 1)}^{2}}{4}}} + c .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18
	$∫(dx/({(x+1)^2 +2^2 }^2 )) x+1=2tanα dx=2sec^2 αdα ∫((2sec^2 αdα)/({2^2 tan^2 α+2^2 }^2 )) ∫((2sec^2 αdα)/({2^2 (1+tan^2 α)}^2 )) ∫((2sec^2 αdα)/(2^4 sec^4 α)) (1/8)∫(dα/(sec^2 α)) (1/(16))∫(1+co2α)dα (1/(16))[α+((sin2α)/2)]+c (1/(16))[tan^(−1) (((x+1)/2))+(1/2)×((2tanα)/(1+tan^2 α))] (1/(16))[tan^(−1) (((x+1)/2))+(((x+1)/2)/(1+(((x+1)/2))^2 ))]+c$
	$\int \frac{d x}{{{(x + 1)}^{2} + 2^{2}}^{2}}$ $x + 1 = 2 t a n α$ $d x = 2 {s e c}^{2} α d α$ $\int \frac{2 {s e c}^{2} α d α}{{2^{2} {t a n}^{2} α + 2^{2}}^{2}}$ $\int \frac{2 {s e c}^{2} α d α}{{2^{2} (1 + {t a n}^{2} α)}^{2}}$ $\int \frac{2 {s e c}^{2} α d α}{2^{4} {s e c}^{4} α}$ $\frac{1}{8} \int \frac{d α}{{s e c}^{2} α}$ $\frac{1}{16} \int (1 + c o 2 α) d α$ $\frac{1}{16} [α + \frac{s i n 2 α}{2}] + c$ $\frac{1}{16} [{t a n}^{- 1} (\frac{x + 1}{2}) + \frac{1}{2} \times \frac{2 t a n α}{1 + {t a n}^{2} α}]$ $\frac{1}{16} [{t a n}^{- 1} (\frac{x + 1}{2}) + \frac{\frac{x + 1}{2}}{1 + {(\frac{x + 1}{2})}^{2}}] + c$
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