3/(x-1)² - 3/(x+1)²
= [3(x+1)²-3(x-1)²]/[(x-1)²(x+1)²]
= (3x²+6x+3-3x²+6x-3)/[(x-1)²(x+1)²]
= (12x)/[(x-1)²(x+1)²]
= (12x)/[(x-1)(x+1)²]
= (12x)/[(x²-1)²]

et donc f(x) = 3/(x-1)² - 3/(x+1)²
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En prenant S pour le signe intégral:

S f(x).dx = 3.S dx/(x-1)² - 3.S dx/(x+1)²
S f(x).dx = -[3/(x-1)]  +  [3/(x+1)]  + C
S f(x).dx = -[3(x+1)-3(x-1)/(x²-1)]  + C
S f(x).dx = [-6/(x²-1)]  + C  
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Remarque un autre manière de faire est de procéder au changement de variable
x² - 1 = t

12x / (x²-1)²
Poser x² - 1 = t
2x.dx = dt

S (12x/(x²-1)²)dx = 6.S dt/t² = -6/t + C
= [-6/(x²-1)] + C
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Sauf distraction