10.7 1) Z
(x+ 3)3dx= Z
(x+ 3)3
| {z }
f3
· 1
|{z}
f′
dx= 14(x+ 3)4+c
2) Z
(2x−1)2dx= Z
(2x−1)2
| {z }
f2
· 2
|{z}
f′
·
1
2dx= 12 Z
(2x−1)2
| {z }
f2
· 2
|{z}
f′
dx
= 12 ·
1
3(2x−1)3 = 16(2x−1)3+c
3) Z
(7x−2)5dx= Z
(7x−2)5
| {z }
f5
· 7
|{z}
f′
·
1
7dx= 17
Z
(7x−2)5
| {z }
f5
· 7
|{z}
f′
dx
= 17 ·
1
6(7x−2)6 = 421 (7x−2)6+c 4)
Z
(3x+ 2)6dx= Z
(3x+ 2)6
| {z }
f6
· 3
|{z}
f′
·
1
3dx= 13 Z
(3x+ 2)6
| {z }
f6
· 3
|{z}
f′
dx
= 13 ·
1
7(3x+ 2)7 = 211 (3x+ 2)7+c 5)
Z
(3x2+x)3(6x+ 1)dx= Z
(3x2+x)3
| {z }
f3
(3x2+x)′
| {z }
f′
dx= 1
4(3x2+x)4+c
6) Z
(4x2−5x)2(16x−10)dx= Z
(4x2−5x)2(8x−5)·2dx
= 2 Z
(4x2−5x)2(8x−5)dx
= 2 Z
(4x2−5x)2(4x2−5x)′dx
= 2·
1
3(4x2−5x)3 = 23 (4x2−5x)3+c
7) Z
x(4x2+ 3)4dx= Z
(4x2+ 3)48x·
1
8dx= 18
Z
(4x2+ 3)48x dx
= 18 Z
(4x2+ 3)4(4x2+ 3)′dx= 18 ·
1
5 (4x2+ 3)5
= 401 (4x2+ 3)5 +c
8) Z
(x2+ 2x) (x3+ 3x2−5)2dx= Z
(x3+ 3x2−5)2(x2+ 2x)·3·
1 3dx
= 13
Z
(x3+ 3x2 −5)2(3x2+ 6x)dx
= 13 Z
(x3+ 3x2 −5)2(x3+ 3x2−5)′dx
= 13·
1
3(x3+3x2−5)3 = 19(x3+3x2−5)3+c
Analyse : primitives Corrigé 10.7
9)
Z 2x+ 1
(x2+x+ 3)2 dx=
Z 1
(x2+x+ 3)2 ·(2x+ 1)dx
= Z
(x2 +x+ 3)−2·(x2+x+ 3)′dx
= −11 (x2+x+ 3)−1 =− 1
x2 +x+ 3+c
10)
Z 3x2
(1 + 2x3)2 dx=
Z 1
(1 + 2x3)2 ·3x2·2·
1 2dx
= 12
Z 1
(1 + 2x3)2 ·6x2dx
= 12 Z
(1 + 2x3)−2·(1 + 2x3)′dx
= 12 ·
1
−1 (1 + 2x3)−1 =− 1
2 (1 + 2x3)+c
11) Z
(3x2+ 1)√
x3+x+ 2dx= Z
(x3+x+ 2)12 (3x2+ 1)dx
= Z
(x3+x+ 2)12 (x3 +x+ 2)′dx
= 1
3 2
(x3+x+ 2)32 = 23p
(x3+x+ 2)3
= 23(x3+x+ 2)√
x3+x+ 2 +c
12) Z
(2x−5)√
x2−5x+ 6dx= Z
(x2−5x+ 6)12 (2x−5)dx
= Z
(x2−5x+ 6)12 (x2−5x+ 6)′dx
= 1
3 2
(x2−5x+ 6)32 = 23
p(x2 −5x+ 6)3
= 23(x2−5x+ 6)√
x2−5x+ 6 +c
13)
Z 1
√3x+ 1dx= Z
(3x+ 1)−12 dx= Z
(3x+ 1)−12 ·3·
1 3dx
= 13 Z
(3x+ 1)12 (3x+ 1)′dx= 13 · 1
1 2
(3x+ 1)12
= 13 ·2√
3x+ 1 = 23
√3x+ 1 +c
14)
Z x+ 1
√x2+ 2x dx=
Z 1
√x2+ 2x(x+ 1)dx
= Z
(x2+ 2x)−12 (x+ 1)·2·
1 2 dx
= 12 Z
(x2+ 2x)−12 (2x+ 2)dx
Analyse : primitives Corrigé 10.7
= 12 Z
(x2+ 2x)−12 (x2+ 2x)′dx= 12 · 1
1 2
(x2+ 2x)12
=√x2+ 2x+c
15)
Z 3x2
√9 +x3 dx=
Z 1
9 +x3 ·3x2dx= Z
(9 +x3)−21 ·3x2dx
= Z
(9+x3)−12 (9+x3)′dx= 1
1 2
(9+x3)21 = 2√
9 +x3+c
16)
Z 3x2
√5x3+ 8dx=
Z 1
√5x3+ 8·3x2·5·
1
5dx = 15
Z
(5x3+ 8)−12·15x2dx
= 15 Z
(5x3+ 8)−21 (5x3+ 8)′dx= 15 · 1
1 2
(5x3+ 8)12
= 25
√5x3+ 8 +c
17) Z
cos(x)p
sin(x)dx= Z
sin(x)12
cos(x)dx= Z
sin(x)12
sin(x)′dx
= 1
3 2
sin(x)32
= 23
q
sin(x)3
= 23 sin(x)p
sin(x) +c
18) Z
sin(x) cos4(x)dx= Z
cos4(x) −sin(x)
·(−1)dx
= (−1) Z
cos4(x) cos(x)′dx=−
1
5 cos5(x) +c 19) cos(x2)′
=−sin(x2)·(x2)′ =−
1
2 sin(x2) Z
cos2(x2) sin(x2)dx= Z
cos2(x2) −
1
2 sin(x2)
·(−2)dx
=−2 Z
cos2(x2) cos(x2)′dx=−2·
1
3 cos3(x2)
=−
2
3 cos3(x2) +c 20)
Z
cos(x)−sin2(x) cos(x)dx= Z
cos(x)dx− Z
sin2(x) cos(x)dx
= Z
cos(x)dx− Z
sin2(x) sin(x)′dx
= sin(x)−
1
3 sin3(x) +c
Analyse : primitives Corrigé 10.7
21)
Z sin(x)
1 + cos(x)2 dx=
Z 1
1 + cos(x)2 sin(x)dx
= Z
1 + cos(x)−2
−sin(x)
·(−1)dx
= (−1) Z
1 + cos(x)−2
1 + cos(x)′dx
= (−1)·
1
−1 1 + cos(x)−1
= 1
1 + cos(x) +c
22)
Z cos(x)
4 sin(x)−13 dx=
Z 1
4 sin(x)−13 cos(x)dx
= Z
4 sin(x)−1−3
·4 cos(x)·
1 4dx
= 14
Z
4 sin(x)−1−3
4 sin(x)−1′dx
= 14 ·
1
−2 4 sin(x)−1−2
=−
1
8 4 sin(x)−12 +c
Analyse : primitives Corrigé 10.7
