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

Za dx =ax +c
Zx dx =x2
2+c
Rxmdx =xm+1
m+ 1 +c m ∈N
Zdx
x2=−1 x+c x 6= 0
Zdx
2√x=√x+c x 6= 0
Zxαdx =xα+1
α+ 1 +c α ∈Q−(−1) et x 6= 0 si α < 0
Zdx
x= ln |x|+c x 6= 0
Zexdx =ex+c
Zaxdx =ax
ln a+c a > 0et a 6= 1
Zsin x dx =−cos x+c
Zcos x dx = sin x+c
Zdx
cos2x=Z(1 + tan2x)dx = tan x+c x 6=π 2+kπ k ∈Z
Zdx
sin2x=−x+c x 6=kπ k ∈Z
Zx dx =x+c
Zx dx =x+c
Zx dx = ln( x) + c
Zdx
2x=Z(1 −2x)dx =x+c
Zcos2x dx =x
2+sin 2x 4+c
Zsin2x dx =x
2−sin 2x 4+c
Ztan x dx =−ln |cos x|+c x 6=π
2+kπ et k ∈Z
Zln x dx =xln x−x+c x > 0

Zdx
sin x= ln |tan(x 2)|+c x 6= (2k+ 1)π et k ∈Z
Zdx
cos x= ln |tan(x 2+π 4)|+c x 6=π 2+ 2kπ et k ∈Z
Zdx
1−x2=1 2ln 
1 + x
1−x
+c|x| 6= 1 x∈]−1,1[
Zdx
√1−x2==−+c|x|<1
Zdx
1 + x2=x+c
Zdx
√x2+ 1 = ln |x+√x2+ 1|+c
Zdx
√x2−1= ln |x+√x2−1|+c|x|>1x > 1)
Zdx
√x2+h= ln |x+√x2+h|+c x2+h > 0
Zdx
x= ln |(x 2)|+c x > 0ou x < 0
Zdx
x= 2 (ex) + c
Zdx
x= ln |x|+c x > 0ou x < 0
Zu(x)v0(x)dx =u(x)v(x)−Zu0(x)v(x)dx
Zb
a
u(x)v0(x)dx = [ u(x)v(x) ]b
a−Zb
a
u0(x)v(x)dx
a6= 0
Zf0(ax +b)dx =1
af(ax +b) + c
Z(ax +b)αdx =1
a
(ax +b)α+1
α+ 1 +c α 6=−1 (ax +b6= 0 si α < 0)
Zdx
ax +b=1 aln |ax +b|+c ax +b6= 0
Zf0[u(x)] u0(x)dx =f[u(x)] + c
Z[u(x)]αu0(x)dx =[u(x)]α+1
α+ 1 +c α 6=−1 (u(x)6= 0 si α < 0)
Zu0(x)
u(x)dx = ln |u(x)|+c u(x)6= 0
Zb
a
f(x)dx =Ztb
ta
f[φ(t)] φ0(t)dt φ
[ta, tb]a=φ(ta), b =φ(tb)