∫((x^4 +x^2 +1)/((x^2 +4)^2 (x^2 +1))) dx


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Question Number 118819 by bramlexs22 last updated on 20/Oct/20

$\int \frac{x^{4} + x^{2} + 1}{{(x^{2} + 4)}^{2} (x^{2} + 1)} d x$
	Answered by MJS_new last updated on 20/Oct/20

	$= \int \frac{13 (x^{2} - 4)}{24 {(x^{2} + 4)}^{2}} + \frac{25}{72 (x^{2} + 4)} + \frac{1}{9 (x^{2} + 1)} d x =$ $= - \frac{13 x}{24 (x^{2} + 4)} + \frac{25}{144} \arctan \frac{x}{2} + \frac{1}{9} \arctan x + C$
	Answered by bobhans last updated on 20/Oct/20
	$in other way by partial fraction ((x^4 +x^2 +1)/((x^2 +4)^2 (x^2 +1))) = ((ax+b)/(x^2 +1)) + ((cx+d)/(x^2 +4)) + ((ex+f)/((x^2 +4)^2 )) x^4 +x^2 +1 = (ax+b)(x^2 +4)^2 +(cx+d)(x^2 +1)(x^2 +4)+(ex+f)(x^2 +1) replacing x by −x yields x^4 +x^2 +1=(−ax+b)(x^2 +4)^2 +(−cx+d)(x^2 +1)(x^2 +4)+(−ex+f)(x^2 +1) from substracting these last two equation gives a=c=e=0 , d=(8/9), f=−((13)/3) therefore I=(1/9)∫(dx/(x^2 +1))+(4/9)∫((2 dx)/(x^2 +4))−((13)/3)∫ (dx/((x^2 +4)^2 )) I= (1/9) arc tan x +(4/9) arc tan ((x/2))−((13)/(48)) (arc tan ((x/2))+((2x)/(x^2 +4))) + C$
	$i n o t h e r w a y b y p a r t i a l f r a c t i o n$ $\frac{x^{4} + x^{2} + 1}{{(x^{2} + 4)}^{2} (x^{2} + 1)} = \frac{a x + b}{x^{2} + 1} + \frac{c x + d}{x^{2} + 4} + \frac{e x + f}{{(x^{2} + 4)}^{2}}$ $x^{4} + x^{2} + 1 = (a x + b) {(x^{2} + 4)}^{2} + (c x + d) (x^{2} + 1) (x^{2} + 4) + (e x + f) (x^{2} + 1)$ $r e p l a c i n g x b y - x y i e l d s$ $x^{4} + x^{2} + 1 = (- a x + b) {(x^{2} + 4)}^{2} + (- c x + d) (x^{2} + 1) (x^{2} + 4) + (- e x + f) (x^{2} + 1)$ $f r o m s u b s t r a c t i n g t h e s e l a s t t w o e q u a t i o n$ $g i v e s a = c = e = 0, d = \frac{8}{9}, f = - \frac{13}{3}$ $t h e r e f o r e I = \frac{1}{9} \int \frac{d x}{x^{2} + 1} + \frac{4}{9} \int \frac{2 d x}{x^{2} + 4} - \frac{13}{3} \int \frac{d x}{{(x^{2} + 4)}^{2}}$ $I = \frac{1}{9} arc \tan x + \frac{4}{9} arc \tan (\frac{x}{2}) - \frac{13}{48} (arc \tan (\frac{x}{2}) + \frac{2 x}{x^{2} + 4}) + C$
	Answered by 1549442205PVT last updated on 21/Oct/20
	$((x^4 +x^2 +1)/((x^2 +4)^2 (x^2 +1))) =((ax^3 +bx^2 +cx+d)/(x^4 +8x^2 +16))+((ex+f)/(x^2 +1)) ⇔x^4 +x^2 +1=(a+e)x^5 +(b+f)x^4 +(a+c+8e)x^3 +(b+d+8f)x^2 +(c+16e)x+d+16f ⇔ { ((a+e=0)),((b+f=1)),((a+c+8e=0)),((b+d+8f=1)),((c+16e=0)),((d+16f=1)) :}⇔ { ((a=c=e=0)),((b=8/9)),((d=−7/9)),((f=1/9)) :} F=∫(((x^4 +x^2 +1)dx)/((x^2 +4)^2 (x^2 +1))) =∫(((8x^2 −7)/(9(x^2 +4)^2 ))+(1/(9(x^2 +1))))dx =(8/9)∫((((x^2 +4)−11)/((x^2 +4)^2 ))+(1/(9(x^2 +1))))dx =(8/9)∫(dx/(x^2 +4))−∫((88dx)/(9(x^2 +4)^2 ))+∫(dx/(9(x^2 +1))) =(4/9)tan^(−1) ((x/2))+(1/9)tan^(−1) (x)−((88)/9)A A=(1/(2.4)).(x/(x^2 +4))+(1/4).(1/2)∫(dx/(x^2 +4)) =(x/(8(x^2 +4)))+(1/8)×(1/2)tan^(−1) (x/2) F=(4/9)tan^(−1) ((x/2))+(1/9)tan^(−1) (x)−((88)/9)[ (1/8)((x/(x^2 +4))+(1/2)tan^(−1) ((x/2)))] =((−1)/6)tan^(−1) ((x/2))+(1/9)tan^(−1) (x) −((11)/9).(x/(x^2 +4))+C$
	$\frac{x^{4} + x^{2} + 1}{{(x^{2} + 4)}^{2} (x^{2} + 1)} = \frac{{ax}^{3} + {bx}^{2} + cx + d}{x^{4} + {8 x}^{2} + 16} + \frac{ex + f}{x^{2} + 1}$ $\Leftrightarrow x^{4} + x^{2} + 1 = (a + e) x^{5} + (b + f) x^{4} + (a + c + 8 e) x^{3}$ $+ (b + d + 8 f) x^{2} + (c + 16 e) x + d + 16 f$ $\Leftrightarrow {\begin{cases} a + e = 0 \\ b + f = 1 \\ a + c + 8 e = 0 \\ b + d + 8 f = 1 \\ c + 16 e = 0 \\ d + 16 f = 1 \end{cases} \Leftrightarrow {\begin{cases} a = c = e = 0 \\ b = 8 / 9 \\ d = - 7 / 9 \\ f = 1 / 9 \end{cases}$ $F = \int \frac{(x^{4} + x^{2} + 1) dx}{{(x^{2} + 4)}^{2} (x^{2} + 1)} = \int (\frac{{8 x}^{2} - 7}{9 {(x^{2} + 4)}^{2}} + \frac{1}{9 (x^{2} + 1)}) dx$ $= \frac{8}{9} \int (\frac{(x^{2} + 4) - 11}{{(x^{2} + 4)}^{2}} + \frac{1}{9 (x^{2} + 1)}) dx$ $= \frac{8}{9} \int \frac{dx}{x^{2} + 4} - \int \frac{88 dx}{9 {(x^{2} + 4)}^{2}} + \int \frac{dx}{9 (x^{2} + 1)}$ $= \frac{4}{9} \tan^{- 1} (\frac{x}{2}) + \frac{1}{9} \tan^{- 1} (x) - \frac{88}{9} A$ $A = \frac{1}{2.4} . \frac{x}{x^{2} + 4} + \frac{1}{4} . \frac{1}{2} \int \frac{dx}{x^{2} + 4}$ $= \frac{x}{8 (x^{2} + 4)} + \frac{1}{8} \times \frac{1}{2} \tan^{- 1} \frac{x}{2}$ $F = \frac{4}{9} \tan^{- 1} (\frac{x}{2}) + \frac{1}{9} \tan^{- 1} (x) - \frac{88}{9} [$ $\frac{1}{8} (\frac{x}{x^{2} + 4} + \frac{1}{2} \tan^{- 1} (\frac{x}{2}))]$ $= \frac{- 1}{6} \tan^{- 1} (\frac{x}{2}) + \frac{1}{9} \tan^{- 1} (x)$ $- \frac{11}{9} . \frac{x}{x^{2} + 4} + C$
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