∫1/(1+x^5)dx
short question , long ans
Help!!!!!!
Hypergeometric Series.
But I do not think this proves that is an elementary function.
Evaluate ∫(1/(1+x^5))dx.
Note that –1 = –1(cos0°+i sin0°)
∴the roots of the equation x^5+1 = 0 are
(–1)^(1/5) (cos((0°+360k°)/5)+i sin((0°+360k°)/5)) , where k = 0, 1, 2, 3, 4
= –1(cos0°+i sin0°) , –1(cos72°+i sin72°) , –1(cos144°+i sin144°) , –1(cos216°+i sin216°) , –1(cos288°+i sin288°)
= –1 , –cos72°–i sin72° , cos36°–i sin36° , cos36°+i sin36° , –cos72°+i sin72°
∴x^5+1
= (x+1)(x+cos72°+i sin72°)(x–cos 36°+i sin36°)(x–cos 36°–i sin36°)(x+cos 72°–i sin72°)
= (x+1)((x–cos36°)²+sin²36°)((x+cos72°)²+sin²72°)
= (x+1)(x²–2x cos36°+cos²36°+sin²36°)(x²+2x cos72°+cos²72°+sin²72°)
= (x+1)(x²–2x sin54°+1)(x²+2x sin18°+1)
For finding the EXACT value of sin18° and sin54°,
Consider 3×18° = 90°–2×18°
sin(3×18°) = sin(90°–2×18°)
sin(3×18°) = cos(2×18°)
3sin18°–4 sin³18° = 1–2 sin²18°
4sin³18°–2sin²18°–3sin18°+1 = 0
(sin18°–1)(4sin²18°+2sin18°–1) = 0
sin18° = 1(rej.) or 4sin²18°+2sin18°–1 = 0
sin18° = (–2±√(2²–4(4)(–1)))/(2×4)
= (–2±√20)/8
= (√5–1)/4 or (–√5–1)/4(rej.)
∴sin54°
= sin(3×18°)
= 3sin18°–4sin³18°
= (3(√5–1))/4–4((√5–1)/4)³
= (3(√5–1))/4–(4(5√5–15+3√5–1))/64
= (12√5–12–5√5+15–3√5+1)/16
= (√5+1)/4
∴(x+1)(x²–2x sin54°+1)(x²+2x sin18°+1)
= (x+1)(x²–((√5+1)x)/2+1)(x²+((√5–1)x)/2+1)
∴Let 1/(1+x^5) ≡ A/(x+1)+(Bx+C)/(x²–((√5+1)x)/2+1)+(Dx+E)(x²+((√5–1)x)/2+1)
1 ≡ A(x²–((√5+1)x)/2+1)(x²+((√5–1)x)/2+1)+(Bx+C)(x+1)(x²+((√5–1)x)/2+1)+(Dx+E)(x+1)(x²–((√5+1)x)/2+1)
1 ≡ A(x^4–x³+x²–x+1)+(Bx+C)(x+1)(x²+((√5–1)x)/2+1)+(Dx+E)(x+1)(x²–((√5+1)x)/2+1)
Put x = –1,
1 = 5A
A = 1/5
∴1 ≡ (x^4–x³+x²–x+1)/5+(Bx+C)(x+1)(x²+((√5–1)x)/2+1)+(Dx+E)(x+1)(x²–((√5+1)x)/2+1
1 ≡ (x^4–x³+x²–x+1)/5+(Bx+C)(x³+((√5+1)x²)/2+((√5+1)x)/2+1)+(Dx+E)(x³–((√5–1)x²)/2–((√5–1)x)/2+1)
1 ≡ (1/5+B+D)x^4+((–2/5+(√5+1)B+2C–(√5–1)D+2E)x³)/2+(2/5+(√5+1)B+(√5+1)C–(√5–1)D–(√5–1)E)x²)/2+((–2/5+2B+(√5+1)C+2D–(√5–1)E)x)/2+1/5+C+E
╭
│1/5+B+D = 0…………………………………………...(1)
│(–2/5+(√5+1)B+2C–(√5–1)D+2E)/2 = 0……………....(2)
∴─┤(2/5+(√5+1)B+(√5+1)C–(√5–1)D–(√5–1)E)/2 = 0……(3)
│(–2/5+2B+(√5+1)C+2D–(√5–1)E = 0…………………(4)
│1/5+C+E = 1…………………………………………...(5)
╰
(3)×2–(2)×2:4/5+(√5–1)C–(√5+1)E = 0……(6)
(3)×2–(4)×2:4/5+(√5–1)B–(√5+1)D = 0……(7)
(7)–(1)×4: (√5–5)B–(√5+5)D = 0……(8)
From (1), B+D = –1/5……(9)
(9)×(√5–5)–(8):2√5D = –(√5–5)/5
D = (5–√5)/(10√5)
= (√5–1)/10
(8)+(9)×(√5+5):2√5B = –(√5+5)/5
B = –(√5+5)/(10√5)
= –(√5+1)/10
(6)–(5)×4: (√5–5)C–(√5+5)E = –4……(10)
From (5), C+E = 4/5……(11)
(11)×(√5–5)–(10):2√5E = –4(√5–5)/5+4 = (40–4√5)/5
E = (40–4√5)/(10√5)
= (4√5–2)/5
(10)+(11)×(√5+5):2√5C = –4–4(√5+5)/5 = –(4√5+40)/5
C = –(4√5+40)/(10√5)
= –(4√5+2)/5
∴∫(1/(1+x^5))dx
= ∫[(1/5)/(x+1)+((–(√5+1)x)/10–(4√5+2)/5)/(x²–((√5+1)x)/2+1)+(((√5–1)x)/10+(4√5–2)/5)/(x²+((√5–1)x)/2+1)]dx
= (ln│x+1│)/5–((√5+1)/10)∫[(x+(8√5+4)/(√5+1))/(x²–((√5+1)x)/2+1)]dx+((√5–1)/10)∫[(x+(8√5–4)/(√5–1))/(x²+((√5–1)x)/2+1)]dx+C_1
= (ln│x+1│)/5+((√5–1)/20)∫[(2x+2√5+18)/(x²+((√5–1)x)/2+1)]dx–((√5+1)/20)∫[(2x–2√5+18)/(x²–((√5+1)x)/2+1)]dx+C_1
= (ln│x+1│)/5+((√5–1)/20)∫[(2x+√5–1)/(x²+((√5–1)x)/2+1)]dx+(((√5–1)(√5+19))/20)∫[dx/(x²+((√5–1)x)/2+1)]–((√5+1)/20)∫[(2x–(√5+1))/(x²–((√5+1)x)/2+1)]dx+(((√5+1)(√5–19))/20)∫[dx/(x²–((√5+1)x)/2+1)]+C_1
= (ln│x+1│)/5+((√5–1)/20)ln│x²+((√5–1)x)/2+1│–((√5+1)/20)ln│x²–((√5+1)x)/2+1│+((9√5–7)/10)∫[dx/((x+(√5–1)/4)²+1–(√5–1)²/16)]–((9√5+7)/10)∫[dx/((x–(√5+1)/4)²+1–(√5+1)²/16))]+C_2
= (ln│x+1│)/5+((√5–1)/20)ln│2x²+(√5–1)x+2│–((√5+1)/20)ln│2x²–(√5+1)x+2│+((9√5–7)/10)∫[dx/((x+(√5–1)/4)²+(2√5+10)/16)]–((9√5+7)/10)∫[dx/((x–(√5+1)/4)²+(10–2√5)/16)]+C_3
= (ln│x+1│)/5+((√5–1)/20)ln│2x²+(√5–1)x+2│–((√5+1)/20)ln│2x²–(√5+1)x+2│+[(9√5–7)/(10√(2√5+10)/4)]tan^(–1) [(x+(√5–1)/4)/(√(2√5+10)/4)]–[(9√5+7)/(10√(10–2√5)/4)]tan^(–1) [(x–(√5+1)/4)/(√(10–2√5)/4)]+C
= (ln│x+1│)/5+((√5–1)/20)ln│2x²+(√5–1)x+2│–((√5+1)/20)ln│2x²–(√5+1)x+2│+[(18√5–14)/(5√(2√5+10))]tan^(–1) [(4x+√5–1)/√(2√5+10)]–[(18√5+14)/(5√(10–2√5))]tan^(–1) [(4x–√5–1)/√(10–2√5)]+C
I did it in this way...............
This exactly the same result which The Integrator gives.