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Question Number 110285 by mathdave last updated on 28/Aug/20

	Answered by 1549442205PVT last updated on 29/Aug/20
	$we have x^8 +4=(x^4 +2)^2 −2x^4 = (x^4 +(√2) x^2 +2)(x^4 −(√2)x^2 +2) ((8x^5 −16x)/(x^8 +4))=((ax^3 +bx^2 +cx+d)/(x^4 −(√2)x^2 +2))+((a_1 x^3 +b_1 x^2 +c_1 x+d_1 )/(x^4 +(√2)x^2 +2)) =((ax^7 +bx^6 +(c+a(√2))x^5 +(b(√2)+d)x^4 +(2a+c(√2))x^3 +(2b+d(√2))x^2 +2cx+2d)/(x^8 +4)) +((a_1 x^7 +b_1 x^6 +(c_1 −a_1 (√2))x^5 (−b_1 (√2)+d_1 )x^4 +(2a_1 −c_1 (√2)x^3 +(2b_1 −d_1 (√2))x^2 +2c_1 x+2d_1 )/(x^8 +4)) ⇔ { (((a+a_1 )=0 (1))),(((b+b_1 ) =0 (2))),(((a−a_1 )(√2)+c+c_1 =8(3))),(((b−b_1 )(√2)+d+d_1 =0 (4))),((2(a+a_1 )+(c−c_1 )(√2)=0(5))),(((d−d_1 )(√2)+2(b+b_1 )=0(6))),((2(c+c_1 )=−16(7))),((d+d_1 =0 (8))) :} (5)(1)⇒c=c_1 ;(1)(3)(7)⇔a=4(√2) ,a_1 =−4(√2) (1)(5)(6)⇒c=c_1 =−4,(2)(6)(8)⇒d=d_1 =0 Therefore ((8x^5 −16x)/(x^8 +4))=((4(√2) x^3 −4)/(x^4 −(√2)x^2 +2))+((−4(√2)x^3 −4)/(x^4 +(√2)x^2 +2)) ∫((8x^5 −16x)/(x^8 +4))dx=(√2)∫((4x^3 −2(√2))/(x^4 −(√2)x^2 +2))dx−(√2)∫((4 x^3 +2(√2))/(x^4 +(√2)x^2 +2))dx+ =(√2)∫ ((dx^4 −(√2)x^2 +2))/(x^4 −(√2)x^2 +2))−(√2)∫ ((d(x^4 −(√2)x^2 +2))/(x^4 −(√2)x^2 +2))dx =(√2)ln∣x^4 −(√2)x^2 +2∣−(√2)ln∣x^4 +(√2)x^2 +2∣ =(√2)ln∣((x^4 −(√2)x^2 +2)/(x^4 −(√2)x^2 +2))∣$
	$we have x^{8} + 4 = {(x^{4} + 2)}^{2} - {2 x}^{4} =$ $(x^{4} + \sqrt{2} x^{2} + 2) (x^{4} - \sqrt{2} x^{2} + 2)$ $\frac{{8 x}^{5} - 16 x}{x^{8} + 4} = \frac{{ax}^{3} + {bx}^{2} + cx + d}{x^{4} - \sqrt{2} x^{2} + 2} + \frac{a_{1} x^{3} + b_{1} x^{2} + c_{1} x + d_{1}}{x^{4} + \sqrt{2} x^{2} + 2}$ $= \frac{{ax}^{7} + {bx}^{6} + (c + a \sqrt{2}) x^{5} + (b \sqrt{2} + d) x^{4} + (2 a + c \sqrt{2}) x^{3} + (2 b + d \sqrt{2}) x^{2} + 2 cx + 2 d}{x^{8} + 4}$ $+ \frac{a_{1} x^{7} + b_{1} x^{6} + (c_{1} - a_{1} \sqrt{2}) x^{5} (- b_{1} \sqrt{2} + d_{1}) x^{4} + ({2 a}_{1} - c_{1} \sqrt{2} x^{3} + ({2 b}_{1} - d_{1} \sqrt{2}) x^{2} + {2 c}_{1} x + {2 d}_{1}}{x^{8} + 4}$ $\Leftrightarrow {\begin{cases} (a + a_{1}) = 0 (1) \\ (b + b_{1}) = 0 (2) \\ (a - a_{1}) \sqrt{2} + c + c_{1} = 8 (3) \\ (b - b_{1}) \sqrt{2} + d + d_{1} = 0 (4) \\ 2 (a + a_{1}) + (c - c_{1}) \sqrt{2} = 0 (5) \\ (d - d_{1}) \sqrt{2} + 2 (b + b_{1}) = 0 (6) \\ 2 (c + c_{1}) = - 16 (7) \\ d + d_{1} = 0 (8) \end{cases}$ $(5) (1) \Rightarrow c = c_{1}; (1) (3) (7) \Leftrightarrow a = 4 \sqrt{2}, a_{1} = - 4 \sqrt{2}$ $(1) (5) (6) \Rightarrow c = c_{1} = - 4, (2) (6) (8) \Rightarrow d = d_{1} = 0$ $Therefore$ $\frac{{8 x}^{5} - 16 x}{x^{8} + 4} = \frac{4 \sqrt{2} x^{3} - 4}{x^{4} - \sqrt{2} x^{2} + 2} + \frac{- 4 \sqrt{2} x^{3} - 4}{x^{4} + \sqrt{2} x^{2} + 2}$ $\int \frac{{8 x}^{5} - 16 x}{x^{8} + 4} dx = \sqrt{2} \int \frac{{4 x}^{3} - 2 \sqrt{2}}{x^{4} - \sqrt{2} x^{2} + 2} dx - \sqrt{2} \int \frac{4 x^{3} + 2 \sqrt{2}}{x^{4} + \sqrt{2} x^{2} + 2} dx +$ $= \sqrt{2} \int \frac{{dx}^{4} - \sqrt{2} x^{2} + 2)}{x^{4} - \sqrt{2} x^{2} + 2} - \sqrt{2} \int \frac{d (x^{4} - \sqrt{2} x^{2} + 2)}{x^{4} - \sqrt{2} x^{2} + 2} dx$ $= \sqrt{2} \ln ∣ x^{4} - \sqrt{2} x^{2} + 2 ∣ - \sqrt{2} \ln ∣ x^{4} + \sqrt{2} x^{2} + 2 ∣$ $= \sqrt{2} \ln ∣ \frac{x^{4} - \sqrt{2} x^{2} + 2}{x^{4} - \sqrt{2} x^{2} + 2} ∣$
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