Integrale indefinito

[Akillez]

Ciao ragazzi,
Scusate se vi tartasso:

$int((x^3-1)/(4x^3+4x+1))$

Come potrei risolverla?

[Kroldar]

divisione tra polinomi e scomposizione in fratti semplici

[Akillez]

penso che il prof abbia sbagliato il testo perchè derive da questa soluzione

$- (45·((√273/576 - 1/64)^(1/3) - (√273/576 + 1/64)^(1/3))/(8·((1593/32 - 81·√273/32)^(1/3) + (81·√273/32 + 1593/32)^(1/3) - 3)·((27·√91/8 - 81·√3/8)^(1/3) + (27·√91/8 + 81·√3/8)^(1/3))) + ((1593/256 - 81·√273/256)^(1/3) + (81·√273/256 + 1593/256)^(1/3) - 3)/(2·((1593/32 - 81·√273/32)^(1/3) + (81·√273/32 + 1593/32)^(1/3) - 3)·((27·√91/8 - 81·√3/8)^(1/3) + (27·√91/8 + 81·√3/8)^(1/3))) - 5·((81·√273/64 - 729/64)^(1/3) - (81·√273/64 + 729/64)^(1/3))/(2·((1593/32 - 81·√273/32)^(1/3) + (81·√273/32 + 1593/32)^(1/3) - 3)·((27·√91/8 - 81·√3/8)^(1/3) + (27·√91/8 + 81·√3/8)^(1/3))) + 2·((1593/32 - 81·√273/32)^(1/3) + (81·√273/32 + 1593/32)^(1/3) + 3)/(((3186 - 162·√273)^(1/3) + (162·√273 + 3186)^(1/3) - 12)·((27·√91/8 - 81·√3/8)^(1/3) + (27·√91/8 + 81·√3/8)^(1/3))))·ATAN((6·x + (3·√273/8 - 27/8)^(1/3) - (3·√273/8 + 27/8)^(1/3))/((27·√91/8 - 81·√3/8)^(1/3) + (27·√91/8 + 81·√3/8)^(1/3))) + (((81·√273/512 - 729/512)^(1/3) - (81·√273/512 + 729/512)^(1/3))/(6·((1593/32 - 81·√273/32)^(1/3) + (81·√273/32 + 1593/32)^(1/3) - 3)) + 15/(32·((1593/32 - 81·√273/32)^(1/3) + (81·√273/32 + 1593/32)^(1/3) - 3)))·LN(3·x^2 + x·((3·√273/8 - 27/8)^(1/3) - (3·√273/8 + 27/8)^(1/3)) + (59/32 - 3·√273/32)^(1/3) + (3·√273/32 + 59/32)^(1/3) + 1) - (((3·√273/512 - 27/512)^(1/3) - (3·√273/512 + 27/512)^(1/3))/((1593/32 - 81·√273/32)^(1/3) + (81·√273/32 + 1593/32)^(1/3) - 3) + 15/(16·((1593/32 - 81·√273/32)^(1/3) + (81·√273/32 + 1593/32)^(1/3) - 3)))·LN(x - (√273/72 - 1/8)^(1/3) + (√273/72 + 1/8)^(1/3)) + x/4$

Cmq grazie dell'aiuto