(1)10.5 1) Z 1 x2 dx= Z x−2dx= 1 −2+1x−2+1 = 1 −1x−1

10.5 1) Z 1

x² dx= Z

x⁻²dx= ¹

−2+1x⁻²⁺¹ = ¹

−1x⁻¹ =− 1 x +c

2) Z 2

x³ dx= Z

2x⁻³dx = 2·

1

−3+1x⁻³⁺¹= 2·

1

−2x⁻² =− 1 x² +c

3) Z

− 7 x⁵ dx=

Z

−7x⁻⁵dx =−7·

1

−5+1x⁻⁵⁺¹=−7·

1

−4x⁻⁴

= 7 4·

1 x⁴ = 7

4x⁴ +c

4) Z

1+ 1 x²

dx=

Z

(1+x⁻²)dx=x+ ¹

−2+1x⁻²⁺¹ =x+ ¹

−1x⁻¹ =x− 1 x+c

5) Z

4 + 2 x² −

5 x⁴

dx=

Z

(4 + 2x⁻²−5x⁻⁴)dx

= 4x+ 2· 1

−2 + 1x⁻²⁺¹−5· 1

−4 + 1x⁻⁴⁺¹

= 4x+ 2· 1

−1x⁻¹−5· 1

−3x⁻³ = 4x− 2 x + 5

3x³ +c

6) Z

− 4 x⁴ −

1 x³ + 3

x⁵

dx = Z

(−4x⁻⁴ −x⁻³+ 3x⁻⁵)dx

=−4 ¹

−4+1x⁻⁴⁺¹−

1

−3+1x⁻³⁺¹+ 3·

1

−5+1x⁻⁵⁺¹

= ⁴₃x⁻³+¹₂x⁻²−

3

4x⁻⁴ = 4

3x³ + 1 2x² −

3 4x⁴ +c

7)

Z √x dx= Z

x¹² dx= 1

1

2 + 1 x¹²⁺¹ = 1

3 2

x³² = ²₃√

x³ = ²₃x√x+c

8) Z

√3 x dx= Z

x¹³ dx= 1

1

3 + 1x¹³⁺¹ = 1

4 3

x⁴³ = ³₄ √³

x⁴ = ³₄x√³x+c

9) Z 1

√xdx= Z

x⁻¹² dx= 1

−

1

2 + 1x⁻¹²⁺¹ = 1

1 2

x¹² = 2√x+c

10)

Z 1

√3

x² dx=

Z

x⁻²³ dx= 1

−²3 + 1x⁻²³⁺¹ = 1

1 3

x¹³ = 3√³ x+c

11) Z

x√x dx= Z √

x³dx= Z

x³² dx= 1

3

2 + 1x³²⁺¹ = 1

5 2

x⁵² = ²₅√ x⁵

= ²₅x²√x+c

12)

Z √x− 1

√x

dx= Z

(x¹² −x⁻¹²)dx= 1

1

2 + 1 x¹²⁺¹− 1

−¹2 + 1x⁻¹²⁺¹

= 1

3 2

x³²− 1

1 2

x¹² = ²₃√

x³−2√x= ²₃x√x−2√x+c

Analyse : primitives Corrigé 10.5

13) Z

√3 x+ 1

√3x

dx = Z

(x¹³ +x⁻¹³)dx = 1

1

3 + 1x¹³⁺¹ + 1

−¹3 + 1x⁻¹³⁺¹ = 1

4 3

x⁴³ + 1

2 3

x²³ = ³₄ √³

x⁴+ ³₂√³

x² = ³₄ x√³x+³₂ √³ x²+c

14) Z

− 2

√3 x+ 1

√3

x⁴

dx= Z

(−2x⁻¹³ +x⁻⁴³)dx

=−2· 1

−

1

3 + 1x⁻¹³⁺¹+ 1

−

4

3 + 1x⁻⁴³⁺¹

=−2· 1

2 3

x²³ + 1

−

1 3

x⁻¹³ =−3√³ x² −

3

√3 x+c

Analyse : primitives Corrigé 10.5