calculate ∫_1 ^(+∞) (dx/((x^2 +1)^2 (x^2 +4)^2 ))


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Question Number 105230 by mathmax by abdo last updated on 27/Jul/20

$calculate \int_{1}^{+ \infty} \frac{dx}{{(x^{2} + 1)}^{2} {(x^{2} + 4)}^{2}}$
	Answered by 1549442205PVT last updated on 27/Jul/20
	$F=∫[(1/4)((1/x^2 )−(1/(x^2 +4)))]^2 dx 16F=∫[(1/x^4 )+(1/((x^2 +4)^2 ))−(2/(x^2 (x^2 +4)))]dx =∫(dx/x^4 )+∫((1/(2(x^2 +4)))−(1/(2x^2 )))dx+∫ (dx/((x^2 +4)^2 )) =−(1/(3x^3 ))+(1/(2x))+(1/2)×(1/2)tan^(−1) ((x/2))+I_2 =−(1/(3x^3 ))+(1/(2x))+(1/4)tan^(−1) ((x/2))+(1/(2.4)).(x/((x^2 +4)))+(1/4).(1/2).(1/2)tan^(−1) ((x/2)) =−(1/(3x^3 ))+(1/(2x))+(x/(8(x^2 +4)))+(5/(16))tan^(−1) ((x/2)) F=(1/(16)){−(1/(3x^3 ))+(1/(2x))+(x/(8(x^2 +4)))+(5/(16))tan^(−1) ((x/2))} I= ∫_1 ^(+∞) (dx/((x^2 +1)^2 (x^2 +4)^2 ))=F(+∞)−F(1) =(1/(16)){(5/(16))×(𝛑/2)−[((23)/(120))+(5/(16))tan^(−1) ((1/2))]}$
	$F = \int {[\frac{1}{4} (\frac{1}{x^{2}} - \frac{1}{x^{2} + 4})]}^{2} dx$ $16 F = \int [\frac{1}{x^{4}} + \frac{1}{{(x^{2} + 4)}^{2}} - \frac{2}{x^{2} (x^{2} + 4)}] dx$ $= \int \frac{dx}{x^{4}} + \int (\frac{1}{2 (x^{2} + 4)} - \frac{1}{{2 x}^{2}}) dx + \int \frac{dx}{{(x^{2} + 4)}^{2}}$ $= - \frac{1}{{3 x}^{3}} + \frac{1}{2 x} + \frac{1}{2} \times \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + I_{2}$ $= - \frac{1}{{3 x}^{3}} + \frac{1}{2 x} + \frac{1}{4} \tan^{- 1} (\frac{x}{2}) + \frac{1}{2.4} . \frac{x}{(x^{2} + 4)} + \frac{1}{4} . \frac{1}{2} . \frac{1}{2} \tan^{- 1} (\frac{x}{2})$ $= - \frac{1}{{3 x}^{3}} + \frac{1}{2 x} + \frac{x}{8 (x^{2} + 4)} + \frac{5}{16} \tan^{- 1} (\frac{x}{2})$ $F = \frac{1}{16} {- \frac{1}{{3 x}^{3}} + \frac{1}{2 x} + \frac{x}{8 (x^{2} + 4)} + \frac{5}{16} \tan^{- 1} (\frac{x}{2})}$ $I = \int_{1}^{+ \infty} \frac{dx}{{(x^{2} + 1)}^{2} {(x^{2} + 4)}^{2}} = F (+ \infty) - F (1)$ $= \frac{1}{16} {\frac{5}{16} \times \frac{π}{2} - [\frac{23}{120} + \frac{5}{16} \tan^{- 1} (\frac{1}{2})]}$
	Commented by abdomsup last updated on 28/Jul/20

	$t h a n k y o u s i r$
	Commented by 1549442205PVT last updated on 28/Jul/20

	$Thank, you are welcome .$
	Answered by mathmax by abdo last updated on 29/Jul/20
	$I =∫_1 ^(+∞) (dx/((x^2 +1)^2 (x^2 +4)^2 )) we have F(x)= (1/((x^2 +1)^2 (x^2 +4)^2 )) =(1/9)((1/(x^2 +1))−(1/(x^2 +4)))^2 =(1/9){ (1/((x^2 +1)^2 )) −(2/((x^2 +1)(x^2 +4))) +(1/((x^2 +4)^2 ))} ⇒ 9F(x) =(1/((x^2 +1)^2 )) −(2/3)((1/(x^2 +1))−(1/(x^2 +4)))+(1/((x^2 +4)^2 )) =(1/((x^2 +1)^2 ))−(2/(3(x^2 +1))) +(2/(3(x^2 +4))) +(1/((x^2 +4)^2 )) ⇒ 9 ∫_1 ^∞ F(x) =∫_1 ^∞ (dx/((1+x^2 )^2 ))−(2/3)∫_1 ^∞ (dx/(1+x^2 )) +(2/3)∫_1 ^∞ (dx/(x^2 +4)) +∫_1 ^∞ (dx/((x^2 +4)^2 )) we have ∫_1 ^∞ (dx/((1+x^2 )^2 )) =_(x=tant) ∫_(π/4) ^(π/2) (((1+tan^2 t))/((1+tan^2 t)^2 ))dt =∫_(π/4) ^(π/2) cos^2 t dt =(1/2)∫_(π/4) ^(π/2) (1+cos(2t))dt =(π/8) +(1/4)[sin(2t)]_(π/4) ^(π/2) =(π/8) −(1/4) and ∫_1 ^∞ (dx/(1+x^2 )) =[arctanx]_1 ^∞ =(π/2)−(π/4) =(π/4) ∫_1 ^∞ (dx/(x^2 +4)) =_(x=2t) ∫_(1/2) ^∞ ((2dt)/(4(1+t^2 ))) =(1/2)((π/2)−arctan((1/2))) =(π/4) −(1/2)((π/2)−arctan2) =((arctan2)/2) ∫_1 ^∞ (dx/((x^2 +4)^2 )) =_(x=2t) ∫_(1/2) ^∞ ((2dt)/(16(t^2 +1)^2 )) =(1/8) ∫_(1/2) ^∞ (dt/((1+t^2 )^2 )) =_(t=tanu) (1/8) ∫_(arctan((1/2))) ^(π/2) ((1+tan^2 u)/((1+tan^2 u)^2 ))du =(1/8)∫_(arctan((1/2))) ^(π/2) ((1+cos(2u))/2) du =(1/(16))∫_(arctan((1/2))) ^(π/2) (1+cos(2u)du =(1/(16))((π/2)−arctan((1/2)))+(1/(32))[sin(2u)]_(...) ^(...) =((arctan2)/(16)) +(1/(32))(−sin(2arctan((1/2))) so the value of I is known$
	$I = \int_{1}^{+ \infty} \frac{dx}{{(x^{2} + 1)}^{2} {(x^{2} + 4)}^{2}}$ $we have F (x) = \frac{1}{{(x^{2} + 1)}^{2} {(x^{2} + 4)}^{2}} = \frac{1}{9} {(\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + 4})}^{2}$ $= \frac{1}{9} {\frac{1}{{(x^{2} + 1)}^{2}} - \frac{2}{(x^{2} + 1) (x^{2} + 4)} + \frac{1}{{(x^{2} + 4)}^{2}}} \Rightarrow$ $9 F (x) = \frac{1}{{(x^{2} + 1)}^{2}} - \frac{2}{3} (\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + 4}) + \frac{1}{{(x^{2} + 4)}^{2}}$ $= \frac{1}{{(x^{2} + 1)}^{2}} - \frac{2}{3 (x^{2} + 1)} + \frac{2}{3 (x^{2} + 4)} + \frac{1}{{(x^{2} + 4)}^{2}} \Rightarrow$ $9 \int_{1}^{\infty} F (x) = \int_{1}^{\infty} \frac{dx}{{(1 + x^{2})}^{2}} - \frac{2}{3} \int_{1}^{\infty} \frac{dx}{1 + x^{2}} + \frac{2}{3} \int_{1}^{\infty} \frac{dx}{x^{2} + 4} + \int_{1}^{\infty} \frac{dx}{{(x^{2} + 4)}^{2}}$ $we have \int_{1}^{\infty} \frac{dx}{{(1 + x^{2})}^{2}} =_{x = tant} \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{(1 + \tan^{2} t)}{{(1 + \tan^{2} t)}^{2}} dt$ $= \int_{\frac{π}{4}}^{\frac{π}{2}} \cos^{2} t dt = \frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} (1 + \cos (2 t)) dt = \frac{π}{8} + \frac{1}{4} {[\sin (2 t)]}_{\frac{π}{4}}^{\frac{π}{2}}$ $= \frac{π}{8} - \frac{1}{4} and \int_{1}^{\infty} \frac{dx}{1 + x^{2}} = {[arctanx]}_{1}^{\infty} = \frac{π}{2} - \frac{π}{4} = \frac{π}{4}$ $\int_{1}^{\infty} \frac{dx}{x^{2} + 4} =_{x = 2 t} \int_{\frac{1}{2}}^{\infty} \frac{2 dt}{4 (1 + t^{2})} = \frac{1}{2} (\frac{π}{2} - \arctan (\frac{1}{2}))$ $= \frac{π}{4} - \frac{1}{2} (\frac{π}{2} - \arctan 2) = \frac{\arctan 2}{2}$ $\int_{1}^{\infty} \frac{dx}{{(x^{2} + 4)}^{2}} =_{x = 2 t} \int_{\frac{1}{2}}^{\infty} \frac{2 dt}{16 {(t^{2} + 1)}^{2}} = \frac{1}{8} \int_{\frac{1}{2}}^{\infty} \frac{dt}{{(1 + t^{2})}^{2}}$ $=_{t = tanu} \frac{1}{8} \int_{\arctan (\frac{1}{2})}^{\frac{π}{2}} \frac{1 + \tan^{2} u}{{(1 + \tan^{2} u)}^{2}} du$ $= \frac{1}{8} \int_{\arctan (\frac{1}{2})}^{\frac{π}{2}} \frac{1 + \cos (2 u)}{2} du = \frac{1}{16} \int_{\arctan (\frac{1}{2})}^{\frac{π}{2}} (1 + \cos (2 u) du$ $= \frac{1}{16} (\frac{π}{2} - \arctan (\frac{1}{2})) + \frac{1}{32} {[\sin (2 u)]}_{. . .}^{. . .}$ $= \frac{\arctan 2}{16} + \frac{1}{32} (- \sin (2 \arctan (\frac{1}{2})) so the value of I is known$
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