Major math poblem - VERY COMPLICATED -

[Bonehelm]

So me and my friends are trying to figure this one out-- I know this isn't something a 16 year old should be attempting...but hey, its summer and I'm bored...


so here it goes...

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 hope you can still breathe..

[PSmith08]

That's nice, but you're still flirting with the boundaries of the real deal.

Prove that the alternating group is a normal subgroup of the symmetric group, and then get back to me.

[Bonehelm]

WTF...why are there smilies again? My formula...it's ruined !!!!

[Wendell_E]

WTF...why are there smilies again? My formula...it's ruined !!!! :-\

Just go down to "Additional Options" and check the "don't use smileys" option:

So me and my friends are trying to figure this one out-- I know this isn't something a 16 year old should be attempting...but hey, its summer and I'm bored...


so here it goes...

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 hope you can still breathe..

[Mark]

You said it yourself: it's summer. Get outside and meet girls. Or boys, if that's your preference.

[Bonehelm]

You said it yourself: it's summer. Get outside and meet girls. Or boys, if that's your preference.

Lol, thanks for the advice, Mark. By the way, from all your observations and posts on this forum, I find it hard to believe that haven't taken any musical education..

[Larry Rinkel]

Lol, thanks for the advice, Mark. By the way, from all your observations and posts on this forum, I find it hard to believe that haven't taken any musical education..

The 8 followed by a paren gives a smiley:

[Mark]

By the way, from all your observations and posts on this forum, I find it hard to believe that haven't taken any musical education..

I learned the art of bluffing! Seriously, though, I have no music education. My secondary (high school) music teacher decided early on that we couldn't be bothered to listen, so he made us copy out stuff from text books and learn precisely nothing, because he never put anything in context or gave examples. It was an utter waste of two years, after which music as a subject was optional ... so many of us opted out. Needless to say, my school quickly got branded a 'failing' establishment by the UK government, and was closed down. It's since reopened in a totally new guise, and I pray for the sake of its pupils today that it's improved.

[PerfectWagnerite]

That integral isn't complicated, just a little tedious in terms of arithmetic. We used to get those as extra credit problems. Teachers like them because they like to brag about how "tough" the problems they give are. What it boils down to is only can you follow a simple procedure. Anything that has a definite procedure to solving it (like problem you stated) isn't complicated. Complicated problems are the ones where you ponder for a few days before you can begin to solve them.

[andy]

That integral isn't complicated, just a little tedious in terms of arithmetic.

Completely agree. This isn't math... it's computation. In other words, this a problem for a computer. Humans are quite likely to make a mistake with this tedious drudgery.

That's nice, but you're still flirting with the boundaries of the real deal.

Prove that the alternating group is a normal subgroup of the symmetric group, and then get back to me.

Indeed, this is real math. Though I'd still agree it's easy

[shive1]

The answer is 2

[aquablob]

So me and my friends are trying to figure this one out-- I know this isn't something a 16 year old should be attempting...but hey, its summer and I'm bored...


so here it goes...

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......(
From (1), B+D = –1/5......(9)
(9)×(?5–5)–(:2?5D = –(?5–5)/5
D = (5–?5)/(10?5)
= (?5–1)/10
(+(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 hope you can still breathe..

42

[Siedler]

simplified version 1+2=3 

[Sarastro]

The answer is...




PS: Now, we should take the derivative  to make sure everything is correct.

[Szykneij]

Yeah, but it took you two years and fourteen days to figure it out.     

[DavidW]

Yeah, but it took you two years and fourteen days to figure it out.     

Or just very bored one night and dug it out.

This is why you don't drink and derive. <worst pun ever>

[Sarastro]

Yeah, but it took you two years and fourteen days to figure it out.     

Well, it took hundreds of years to figure out some mathematical problems, and some are not yet figured out at all.

[Szykneij]

Well, it took hundreds of years to figure out some mathematical problems, and some are not yet figured out at all.

Of course, the fact it took nearly four months to formulate that reply ...

[Coopmv]

Some mathematicians are still working on the next largest prime.  I doubt whoever that discovers the next largest prime will win the Crawford Prize, the Nobel Prize in mathematics ...

[Opus106]

Some mathematicians are still working on the next largest prime.  I doubt whoever that discovers the next largest prime will win the Crawford Prize, the Nobel Prize in mathematics ...

I'm afraid it doesn't work that way, Stuart. These days, just about anyone can help to find the next prime; for example, by participating in a community project like PrimeGrid. All you need to do is spare some computing power of your PC, and you could make the headlines of every major (geek) news outlet!

And what's more -- the list of primes is infinite! The endowment would probably dwindle if people are rewarded for discovering a prime. Although if an insight into the nature of primes itself is offered, then that might be considered worthy of an award.

And finally -- this is probably just a matter of opinion, but usually the Fields Medal is considered the "Nobel Prize" of mathematics.