decompose F(x) =((2x−1)/((x^2 −1)^2 (x^2 +3))) and calculate ∫


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Question Number 119425 by mathmax by abdo last updated on 24/Oct/20

$decompose F (x) = \frac{2 x - 1}{{(x^{2} - 1)}^{2} (x^{2} + 3)}$ $and calculate \int_{\sqrt{2}}^{+ \infty} F (x) dx$
	Answered by 1549442205PVT last updated on 24/Oct/20
	$F(x) =((2x−1)/((x^2 −1)^2 (x^2 +3)))=((ax+b)/(x^2 +3))+((mx^3 +nx^2 +px+q)/(x^4 −2x^2 +1)) ⇔(a+m)x^5 +(b+n)x^4 +(3m+p−2a)x^3 +(3n+q−2b)x^2 +(a+3p)x+b+3q=2x−1 ⇔ { ((a+m=0)),((b+n=0)),((3m+p−2a=0)),((3n+q−2b=0)),((a+3p=2)),((b+3q=−1)) :} { ((5m+p=0)),((5n+q=0)),((−m+3p=2)),((−n+3q=−1)) :} 16q=−5⇒q=−5/16⇒n=1/16 16p=10⇒p=10/16⇒m=−2/16 ⇒a=2/16,b=−1/16 Hence we have F(x) =((2x−1)/((x^2 −1)^2 (x^2 +3))) =((2x−1)/(16(x^2 +3)))−((2x^3 −x^2 −10x+5)/(16(x^2 −1)^2 )) 2x^3 −x^2 −10x+5=(x^2 −1)(2x−1)−8x+4 ⇒((2x^3 −x^2 −10x+5)/(16(x^2 −1)^2 ))=((2x−1)/(16(x^2 −1)))−((8x−4)/(16(x^2 −1)^2 )) (2/(x^2 −1))=(1/(x−1))−(1/(x+1))=a−b⇒2ab=a−b ⇒(4/((x^2 −1)^2 ))=(a−b)^2 =a^2 +b^2 −(a−b) ⇒F(x)=((2x−1)/(16(x^2 +3)))−((2x−1)/(16(x^2 −1)))+((4(2x−1))/(16(x^2 −1)^2 )) ∫_(√2) ^∞ F(x)=∫_(√2) ^∞ ((d(x^2 +3))/(16(x^2 +3))−(1/(16))∫_(√2) ^∞ (dx/((x^2 +3))) −(1/(16))∫_(√2) ^∞ ((d(x^2 −1))/(x^2 −1))+(1/(16))∫_(√2) ^∞ (dx/(x^2 −1))+(1/4)∫_(√2) ^∞ ((d(x^2 −1))/((x^2 −1)^2 )) −(1/4)∫_(√2) ^∞ (dx/((x^2 −1)^2 )) (1).We have J=(1/(16))∫_(√2) ^∞ (dx/((x^2 −1)))=(1/(32))∫_(√2) ^∞ ((1/(x−1))−(1/(x+1)))dx =(1/(32))ln∣((x−1)/(x+1))∣_(√2) ^∞ =−(1/(32))ln(((√2)−1)/( (√2)+1))=−(1/(16))ln((√2)−1) I=(1/4)∫_(√2) ^∞ (dx/((x^2 −1)^2 ))=(1/(16))∫_(√2) ^∞ ((1/(x−1))−(1/(x+1)))^2 dx= (1/(16))∫_(√2) ^∞ ((1/((x−1)^2 ))+(1/((x+1)^2 ))−(1/(x−1))+(1/(x+1)))dx(2) =(1/(16))[[−(1/(x−1))−(1/(x+1))+ln∣((x+1)/(x−1∣))]_(√2) ^∞ =(1/(64))((1/( (√2)−1))+(1/( (√2)+1))−ln(((√2)+1)/( (√2)−1)))=(1/(16))[2(√2) −2ln((√2)+1)] From (1)(2)we get ∫_(√2) ^∞ F(x) dx=∫_(√2) ^∞ ((2x−1)/((x^2 −1)^2 (x^2 +3)))dx =[(1/(16))ln(x^2 +3)−(1/(16)).(1/( (√3)))tan^(−1) ((x/( (√3)))) −(1/(16))ln∣x^2 −1∣−(1/(4(x^2 −1)))]_(√2) ^∞ +J−I =(1/(16))ln((x^2 +3)/(x^2 −1))−(1/(16(√3)))tan^(−1) ((x/( (√3))))−(1/(4(x^2 −1)))]_(√2) ^∞ −(1/(16))ln((√2)−1)−(1/(16))(2(√2)−2ln(1+(√2) ) =−(π/(32(√3)))−(1/(16))ln5+(1/(16(√3)))tan^(−1) (((√6)/3))+(1/4) −((√2)/8)+(3/(16))ln((√2) +1)≈0.1059174283$
	$F (x) = \frac{2 x - 1}{{(x^{2} - 1)}^{2} (x^{2} + 3)} = \frac{ax + b}{x^{2} + 3} + \frac{{mx}^{3} + {nx}^{2} + px + q}{x^{4} - {2 x}^{2} + 1}$ $\Leftrightarrow (a + m) x^{5} + (b + n) x^{4} + (3 m + p - 2 a) x^{3}$ $+ (3 n + q - 2 b) x^{2} + (a + 3 p) x + b + 3 q = 2 x - 1$ $\Leftrightarrow {\begin{cases} a + m = 0 \\ b + n = 0 \\ 3 m + p - 2 a = 0 \\ 3 n + q - 2 b = 0 \\ a + 3 p = 2 \\ b + 3 q = - 1 \end{cases} {\begin{cases} 5 m + p = 0 \\ 5 n + q = 0 \\ - m + 3 p = 2 \\ - n + 3 q = - 1 \end{cases}$ $16 q = - 5 \Rightarrow q = - 5 / 16 \Rightarrow n = 1 / 16$ $16 p = 10 \Rightarrow p = 10 / 16 \Rightarrow m = - 2 / 16$ $\Rightarrow a = 2 / 16, b = - 1 / 16$ $Hence we have$ $F (x) = \frac{2 x - 1}{{(x^{2} - 1)}^{2} (x^{2} + 3)}$ $= \frac{2 x - 1}{16 (x^{2} + 3)} - \frac{{2 x}^{3} - x^{2} - 10 x + 5}{16 {(x^{2} - 1)}^{2}}$ ${2 x}^{3} - x^{2} - 10 x + 5 = (x^{2} - 1) (2 x - 1) - 8 x + 4$ $\Rightarrow \frac{{2 x}^{3} - x^{2} - 10 x + 5}{16 {(x^{2} - 1)}^{2}} = \frac{2 x - 1}{16 (x^{2} - 1)} - \frac{8 x - 4}{16 {(x^{2} - 1)}^{2}}$ $\frac{2}{x^{2} - 1} = \frac{1}{x - 1} - \frac{1}{x + 1} = a - b \Rightarrow 2 ab = a - b$ $\Rightarrow \frac{4}{{(x^{2} - 1)}^{2}} = {(a - b)}^{2} = a^{2} + b^{2} - (a - b)$ $\Rightarrow F (x) = \frac{2 x - 1}{16 (x^{2} + 3)} - \frac{2 x - 1}{16 (x^{2} - 1)} + \frac{4 (2 x - 1)}{16 {(x^{2} - 1)}^{2}}$ $\int_{\sqrt{2}}^{\infty} F (x) = \int_{\sqrt{2}}^{\infty} \frac{d (x^{2} + 3)}{16 (x^{2} + 3} - \frac{1}{16} \int_{\sqrt{2}}^{\infty} \frac{dx}{(x^{2} + 3)}$ $- \frac{1}{16} \int_{\sqrt{2}}^{\infty} \frac{d (x^{2} - 1)}{x^{2} - 1} + \frac{1}{16} \int_{\sqrt{2}}^{\infty} \frac{dx}{x^{2} - 1} + \frac{1}{4} \int_{\sqrt{2}}^{\infty} \frac{d (x^{2} - 1)}{{(x^{2} - 1)}^{2}}$ $- \frac{1}{4} \int_{\sqrt{2}}^{\infty} \frac{dx}{{(x^{2} - 1)}^{2}} (1) . We have$ $J = \frac{1}{16} \int_{\sqrt{2}}^{\infty} \frac{dx}{(x^{2} - 1)} = \frac{1}{32} \int_{\sqrt{2}}^{\infty} (\frac{1}{x - 1} - \frac{1}{x + 1}) dx$ $= \frac{1}{32} \ln ∣ \frac{x - 1}{x + 1} ∣_{\sqrt{2}}^{\infty} = - \frac{1}{32} \ln \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = - \frac{1}{16} \ln (\sqrt{2} - 1)$ $I = \frac{1}{4} \int_{\sqrt{2}}^{\infty} \frac{dx}{{(x^{2} - 1)}^{2}} = \frac{1}{16} \int_{\sqrt{2}}^{\infty} {(\frac{1}{x - 1} - \frac{1}{x + 1})}^{2} dx =$ $\frac{1}{16} \int_{\sqrt{2}}^{\infty} (\frac{1}{{(x - 1)}^{2}} + \frac{1}{{(x + 1)}^{2}} - \frac{1}{x - 1} + \frac{1}{x + 1}) dx (2)$ $= \frac{1}{16} [{[- \frac{1}{x - 1} - \frac{1}{x + 1} + \ln ∣ \frac{x + 1}{x - 1 ∣}]}_{\sqrt{2}}^{\infty}$ $= \frac{1}{64} (\frac{1}{\sqrt{2} - 1} + \frac{1}{\sqrt{2} + 1} - \ln \frac{\sqrt{2} + 1}{\sqrt{2} - 1}) = \frac{1}{16} [2 \sqrt{2} - 2 \ln (\sqrt{2} + 1)]$ $From (1) (2) we get$ $\int_{\sqrt{2}}^{\infty} F (x) dx = \int_{\sqrt{2}}^{\infty} \frac{2 x - 1}{{(x^{2} - 1)}^{2} (x^{2} + 3)} dx$ $= [\frac{1}{16} \ln (x^{2} + 3) - \frac{1}{16} . \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}})$ ${- \frac{1}{16} \ln ∣ x^{2} - 1 ∣ - \frac{1}{4 (x^{2} - 1)}]}_{\sqrt{2}}^{\infty} + J - I$ ${= \frac{1}{16} \ln \frac{x^{2} + 3}{x^{2} - 1} - \frac{1}{16 \sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) - \frac{1}{4 (x^{2} - 1)}]}_{\sqrt{2}}^{\infty}$ $- \frac{1}{16} \ln (\sqrt{2} - 1) - \frac{1}{16} (2 \sqrt{2} - 2 \ln (1 + \sqrt{2})$ $= - \frac{π}{32 \sqrt{3}} - \frac{1}{16} \ln 5 + \frac{1}{16 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{6}}{3}) + \frac{1}{4}$ $- \frac{\sqrt{2}}{8} + \frac{3}{16} \ln (\sqrt{2} + 1) \approx 0.1059174283$
	Commented by Bird last updated on 24/Oct/20

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