Exercice 6. (décomposition en éléments simples)

1
2x2+3x−4, 1 −x2,x(1+2x),
2x+1
x,3x−1 2x,x−3 x,
x√x,√x+1
√x
I1=Z4
2
x3−4x2+x
x2x
I2=Z0
−3
x2−x−2 x
I3=Zπ/3
0
(1−cos(3x)) x
I4=Zπ/2
0
(sin x+cos x)x
I5=Zπ/2
0
sin2x x
I6=Zt
0
tan2x x
I7=Zπ
0
(cos 2x+sin 2x)2x
I8=Z7π/6
π/4
(cos 4x+2 sin xcos x)x
f[0, π/2]
f(x) = sin x
sin x+cos x
a b
x∈[0, π/2]
f(x) = bcos x−sin x
cos x+sin x+a
α∈[0, π/4]α
Zπ/2−α
α
f(x)x.
f
f(x) = 12x2−37x+13
(5x−2)(2x−1)2
f
a b c x
f
f(x) = a
5x−2+bx +c (2x−1)2.
f] −
∞, 2/5[
R0
−1f(x)x
a b x ∈
R\ {−1, 1}
4x−5
x2−1=a x+1+b x−1.
I=Z5
3
4x−5
x2−1x.
x2−x
x+1,2x−1 (x−1)(x2+4),2x+1 (x−2)(x+1)2.
I1=Zπ/6
0
sin 4xcos 2x x
I2=Zπ/2
0
cos 3xsin 5x x
I3=Zπ/4
0
sin4xcos2x x
I1=2Z1
0
(x−1)(x2−2x+2)3x
I2=Zπ/3
π/4
1+tan2x
tan xx
I3=Z2
0
xexp(x2)x
x4
√3+x5,√ln x
x,x2 x3+2
exp(x)
exp(x) + 2, tan x,1 xln x,
a b
I1=Zπ/2
0
sin 2t
a2cos2t+b2sin2tt,
I2=Zπ/3
0
sin x
1−3 cos xx

I1=Z1
0
√x+2x,
I2=Z2
0
x√x+1x,
I3=Z3
1
√3−x x,
I4=Z5
0
√5−x x.
xexp(−x),xexp(2x),xsin x,x4ln x,
x2exp(x), , x33x,xtan2x,xcos2x,
ln x, ln(x2+1),(ln x)3, sin(ln x), cos(ln x).
I=Zπ/2
0
16xcos4x x.
f g R
f(x) = sin2xcos x,g(x) = sin3x.
F f G
g
Zπ/2
0
9xsin2xcos xdx.
I1=Z1
0
ln(x2−2x+4)x,
I2=Z2/3
0
ln(9x2+6x+4)x,
I3=Z4
1
ln(2x2+5x−3)x,
I4=Z−3
−4
arctan x+3
x+5x.
Zx
√4x2+4x+1Zx √x2−4x,
Zx
√x2−4x+9,Zθ √5+2θ−4θ2,
Zϕ
√2+ϕ−ϕ2,Zx √x2+2x+3.
Zπ/3
0
x
cos2(x) + 3 sin2(x)
Z2π/3
0
x
5+4 cos(x)
Zπ/4
0
x
cos3(x)
Zπ/2
0
sin 2x
(2+cos(x))2x
Zπ/2
0
x
cos(x) − 2 sin(x) + 3
f: [a;b]→RF
Rb
af(x)x F(b) − F(a)
f
[a;b]M
x∈[a;b]|f0(x)|6M n ∈N
xk
xk=a+k
n(b−a),k=0, 1, ···n.
x0=a xn=b
Sn=
n−1
X
k=0
f(xk)b−a
n.
F(b) − F(a) =
n−1
X
k=0
(F(xk+1) − F(xk)).
ξk∈]xk;xk+1[k=0, 1, ···n−1
F(b) − F(a) = Sn+
n−1
X
k=0
f0(ξk)
2b−a n2
.
σn
|σn|6M(b−a)2
2n.
limn→+∞Sn
lim
n→+∞
Sn=F(b) − F(a).