FORMULAIRE d`INTÉGRATION Dans ce qui suit "c" est une

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Z
Z
R
Z
Z
Z
Z
Z
a dx = ax + c
x2
+c
x dx =
2
xm+1
xm dx =
+c
m∈N
m+1
dx
1
=− +c
x 6= 0
2
x
x
√
dx
√ = x+c
x 6= 0
2 x
xα+1
α
x dx =
+c
α ∈ Q − (−1) et x 6= 0 si α < 0
α+1
dx
= ln |x| + c
x 6= 0
x
ex dx = ex + c
ax
+c
a > 0 et a 6= 1
ln a
Z
sin x dx = − cos x + c
Z
ax dx =
Z
cos x dx = sin x + c
Z
dx
=
(1 + tan2 x) dx = tan x + c
cos2 x
Z
dx
=−
x+c
x=
6 kπ k ∈ Z
sin2 x
Z
%(;"1gC!
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Z
#/i x dx = %ji x + c
Z
%gi x dx = #/i x + c
Z
1=i x dx = ln(%gi x) + c
Z
Z
dx
%gi = (1 − 1/i x) dx = 1=i
2
2
x
x sin 2x
+
+c
2
4
Z
x sin 2x
sin2 x dx = −
+c
2
4
Z
tan x dx = − ln | cos x| + c
Z
Z
x 6=
π
+ kπ k ∈ Z
2
x+c
cos2 x dx =
ln x dx = x ln x − x + c
π
+ kπ et k ∈ Z
2
x>0
x 6=
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Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
dx
x
= ln | tan( )| + c
x 6= (2k + 1)π et k ∈ Z
sin x
2
dx
x π
π
= ln | tan( + )| + c x 6= + 2kπ et k ∈ Z
cos x
2 4
2
1 + x
1
dx
+c
x ∈] − 1, 1[ =
ln
|x|
=
6
1
1 − x2
2
1−x
dx
√
= −
+c
|x| < 1
=
2
1−x
dx
x +c
=
1 + x2
√
dx
√
=
ln
|x
+
x2 + 1| + c 2
x +1
√
dx
2 − 1| + c
√
x
x > 1)
=
ln
|x
+
|x|
>
1
x2 − 1
√
dx
√
=
ln
|x
+
x2 + h| + c x2 + h > 0
2
x +h
x
dx
= ln | ( )| + c x > 0 ou x < 0
x
2
dx
=2
(ex ) + c
x
dx
= ln | x| + c x > 0 ou x < 0
x
;+
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1=i
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Z
0
u(x) v (x) dx = u(x) v(x) −
Z
Z
u0 (x) v(x)dx
b
a
#%$ &GM
Pc$YZc-RXW/.
u(x) v 0 (x) dx = [ u(x) v(x) ]ba −
Z
+ '(1$- ?7><'7!,1/- "!0@A'(#
b
u0 (x) v(x)dx
a
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a 6= 0 1
f (ax + b) + c
a
Z
1 (ax + b)α+1
(ax + b)α dx =
+ c α 6= −1 (ax + b 6= 0 si α < 0)
a
α+1
Z
1
dx
= ln |ax + b| + c ax + b 6= 0
ax + b
a
Z
f 0 (ax + b) dx =
Pc$Y021R31] ced4.
Z
f 0 [u(x)] u0 (x) dx = f [u(x)] + c
Z
[u(x)]α u0 (x) dx =
Z
[u(x)]α+1
+c
α+1
u0 (x)
dx = ln |u(x)| + c
u(x)
Z
b
f (x)dx =
a
Z
u(x) 6= 0
tb
f [φ(t)] φ0 (t) dt
ta
α 6= −1 (u(x) 6= 0 si α < 0)
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#<+$> [t , t ] 'L1 a = φ(t ) , b = φ(t )
a
b
a
b