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Question Number 96282 by 675480065 last updated on 31/May/20

	Answered by bemath last updated on 31/May/20

	$look qn 94383$
	Answered by 1549442205 last updated on 31/May/20
	$We have x^4 +1=(x^2 +1)^2 −2x^2 =(x^2 +(√2) x+1)(x^2 −(√2) x+1) Hence,(1/(x^4 +1))=((ax+b)/(x^2 +(√(2 ))x+1))+((cx+d)/(x^2 −(√2)x+1)) ⇔1=(a+c)x^3 +[b+d+(√2)(c−a)]x^2 +[(√2)(c−b)+a+c]x+b+d { ((a+c=0)),((b+d=1)) :} { ((b+d+(√2)(c−a)=0)),(((√2)(c−b)+a+c=0)) :} Deduce a=(1/(2(√2))),b=c=((−1)/(2(√2))),d=1+(1/(2(√2))). hen ∫(dx/(x^4 +1))=(1/(2(√2)))∫((x−1)/(x^2 +(√2)x+1))dx−(1/(2(√2)))∫((x−2(√(2−))1)/(x^2 −(√2)x+1))dx= I=∫((x−1)/(x^2 +(√2)x+1))dx=(1/2)∫((d(x^2 +(√2)x+1))/(x^2 +(√2)x+1))dx+(1/2)∫((−(√2)−2)/(x^2 +(√2)x+1))dx =(1/2)∫((d(x^2 +(√2)x+1))/(x^2 +(√2)x+1))dx−((2+(√2))/2)∫(dx/((x+((√2)/2))^2 +((1/(√2)))^2 )) =(1/2)ln(x^2 +(√2)x+1)−((2+(√2))/2)[((√2)x+1)arctan((√2)x+1)] J=∫((x−2(√2)−1)/(x^2 −(√2)x+1))dx=(1/2)∫((d(x^2 −(√2)x+1))/(x^2 −(√2)x+1))+(1/2)∫((−3(√2)−2)/(x^2 −(√2)x+1))dx =(1/2)ln(x^2 −(√2)x+1)−((3(√2)+2)/2)[((√2)x−1)arctan((√2)x−1) From that F(x)= ∫(dx/(x^4 +1))=(1/(4(√2)))ln(x^2 +(√2)x+1)−((2+(√2))/(4(√2)))[((√2)x+1)arctan((√2)x+1)] −(1/(4(√2)))ln(x^2 −(√2)x+1)+((3(√2)+2)/(4(√2)))[((√2)x−1)arctan((√2)x−1)+C F(x)=(1/(4(√2)))ln((x^2 +(√2)x+1)/(x^2 −(√2)x+1))−(((√2)+1)/4)[((√2)x+1)arctan((√2)x+1)]+((3+(√2))/4)[((√2)x−1)arctan((√2)x−1)]+C$
	$We have x^{4} + 1 = {(x^{2} + 1)}^{2} - {2 x}^{2} = (x^{2} + \sqrt{2} x + 1) (x^{2} - \sqrt{2} x + 1)$ $Hence, \frac{1}{x^{4} + 1} = \frac{ax + b}{x^{2} + \sqrt{2} x + 1} + \frac{cx + d}{x^{2} - \sqrt{2} x + 1}$ $\Leftrightarrow 1 = (a + c) x^{3} + [b + d + \sqrt{2} (c - a)] x^{2} + [\sqrt{2} (c - b) + a + c] x + b + d$ ${\begin{cases} a + c = 0 \\ b + d = 1 \end{cases}$ ${\begin{cases} b + d + \sqrt{2} (c - a) = 0 \\ \sqrt{2} (c - b) + a + c = 0 \end{cases}$ $Deduce a = \frac{1}{2 \sqrt{2}}, b = c = \frac{- 1}{2 \sqrt{2}}, d = 1 + \frac{1}{2 \sqrt{2}} .$ $hen \int \frac{dx}{x^{4} + 1} = \frac{1}{2 \sqrt{2}} \int \frac{x - 1}{x^{2} + \sqrt{2} x + 1} dx - \frac{1}{2 \sqrt{2}} \int \frac{x - 2 \sqrt{2 -} 1}{x^{2} - \sqrt{2} x + 1} dx =$ $I = \int \frac{x - 1}{x^{2} + \sqrt{2} x + 1} dx = \frac{1}{2} \int \frac{d (x^{2} + \sqrt{2} x + 1)}{x^{2} + \sqrt{2} x + 1} dx + \frac{1}{2} \int \frac{- \sqrt{2} - 2}{x^{2} + \sqrt{2} x + 1} dx$ $= \frac{1}{2} \int \frac{d (x^{2} + \sqrt{2} x + 1)}{x^{2} + \sqrt{2} x + 1} dx - \frac{2 + \sqrt{2}}{2} \int \frac{dx}{{(x + \frac{\sqrt{2}}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2}}$ $= \frac{1}{2} \ln (x^{2} + \sqrt{2} x + 1) - \frac{2 + \sqrt{2}}{2} [(\sqrt{2} x + 1) \arctan (\sqrt{2} x + 1)]$ $J = \int \frac{x - 2 \sqrt{2} - 1}{x^{2} - \sqrt{2} x + 1} dx = \frac{1}{2} \int \frac{d (x^{2} - \sqrt{2} x + 1)}{x^{2} - \sqrt{2} x + 1} + \frac{1}{2} \int \frac{- 3 \sqrt{2} - 2}{x^{2} - \sqrt{2} x + 1} dx$ $= \frac{1}{2} \ln (x^{2} - \sqrt{2} x + 1) - \frac{3 \sqrt{2} + 2}{2} [(\sqrt{2} x - 1) \arctan (\sqrt{2} x - 1)$ $From that F (x) = \int \frac{dx}{x^{4} + 1} = \frac{1}{4 \sqrt{2}} \ln (x^{2} + \sqrt{2} x + 1) - \frac{2 + \sqrt{2}}{4 \sqrt{2}} [(\sqrt{2} x + 1) \arctan (\sqrt{2} x + 1)]$ $- \frac{1}{4 \sqrt{2}} \ln (x^{2} - \sqrt{2} x + 1) + \frac{3 \sqrt{2} + 2}{4 \sqrt{2}} [(\sqrt{2} x - 1) \arctan (\sqrt{2} x - 1) + C$ $F (x) = \frac{1}{4 \sqrt{2}} \ln \frac{x^{2} + \sqrt{2} x + 1}{x^{2} - \sqrt{2} x + 1} - \frac{\sqrt{2} + 1}{4} [(\sqrt{2} x + 1) \arctan (\sqrt{2} x + 1)] + \frac{3 + \sqrt{2}}{4} [(\sqrt{2} x - 1) \arctan (\sqrt{2} x - 1)] + C$
	Answered by mathmax by abdo last updated on 31/May/20

	$\int \frac{dx}{x^{4} + 1} = \int \frac{\frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} dx = \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}} - (1 - \frac{1}{x^{2}})}{x^{2} + \frac{1}{x^{2}}} dx$ $= \frac{1}{2} \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 2} dx (\to u = x - \frac{1}{x}) - \frac{1}{2} \int \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2} - 2} (\to v = x + \frac{1}{x})$ $= \frac{1}{2} \int \frac{du}{u^{2} + 2} - \frac{1}{2} \int \frac{dv}{v^{2} - 2} but$ $\int \frac{du}{u^{2} + 2} =_{u = \sqrt{2} z} \int \frac{\sqrt{2} dz}{2 (1 + z^{2})} = \frac{1}{\sqrt{2}} \arctan (z) + c_{0} = \frac{1}{\sqrt{2}} \arctan (\frac{u}{\sqrt{2}}) + c_{0}$ $= \frac{1}{\sqrt{2}} \arctan (\frac{1}{\sqrt{2}} (x - \frac{1}{x})) + c_{0}$ $\int \frac{dv}{v^{2} - 2} dv = \int \frac{dv}{(v - \sqrt{2}) (v + \sqrt{2})} = \frac{1}{2 \sqrt{2}} \int (\frac{1}{v - \sqrt{2}} - \frac{1}{v + \sqrt{2}})$ $= \frac{1}{2 \sqrt{2}} \ln ∣ \frac{v - \sqrt{2}}{v + \sqrt{2}} ∣ + c_{1} = \frac{1}{2 \sqrt{2}} \ln ∣ \frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}} ∣ + c_{1} \Rightarrow$ $\int \frac{dx}{x^{4} + 1} = \frac{1}{2 \sqrt{2}} \arctan (\frac{1}{\sqrt{2}} (x - \frac{1}{x})) - \frac{1}{4 \sqrt{2}} \ln ∣ \frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}} ∣ + C$
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