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48
⟹ ∫ 1
1
x 2 − 1 dx = ∫ ⁄ 2
x − 1 + − 1⁄
2
x + 1 dx = 1 2 ∫ 1
x − 1 dx − 1 2 ∫ 1
x + 1 dx
• f(x) = ∫
= 1 2 ln|x − 1| − 1 ln|x + 1|
2
1
x 2 +2x−3
dx ; a = 1; b = 2; c = −3 ; ∆= 16 > 0
x = −b±√b2 −4ac
2a
⟹ x ′ = −3 et x ′′ = 1
Factorisation :
ax 2 + bx + c = a(x − x ′ )(x − x ′′ )
x 2 + 2x − 3 = (x + 3)(x − 1)
Décomposition en élément simple :
1
x 2 + 2x − 3 = 1
(x + 3)(x − 1) =
α
(x + 3) +
β
(x − 1)
α =
1
(x+3) = 1 4
X=1
1
1
x 2 + 2x − 3 = 4
(x + 3) +
−1
4
(x − 1)
1
1
∫ dx = ∫ 4
+
x 2 +2x−3 (x+3)
1 ln|x − 1| − 1 ln|x + 3| + c
4 4
−1
4
(x−1) dx = 1 4 ∫ 1
(x+3) dx − 1 4 ∫ 1
(x+3) dx