Integrale Inmediatas Cambio De Variables

INTEGRALES INMEDIATAS (CAMBIO DE VARIABLES)

Resuelve los siguientes ejercicios de la página 416 del libro de James Stewart.

2. ∫x (4+x2)10 dx

u=x dx

du=2x

1/2 ∫x^10 du = 1/2 [u^11/11 du]+ c = 1/2 [((4+x^2 )^11)/11]+c

((4+x^2 )^11)/22+c

3. ∫x^2 √(x^3 )+ 1 dx

u= x^3+ 1

du=3x^2 dx

du/3= x^2 dx

1/3 ∫u^(1/2 ) du = 1/3 [ u^(3/2)/(3/2) ]du+c = 1/3 [2 u^(3/2)/3 ]+ c = (2u^(3/2 ))/9+ c

2/9 ( x^3+1)^(3/2) +c

5. ∫ 4/((1+2x)^3 ) dx

u=1+2x

du=2 dx

[2/((1+2x)^3 )+2/((1+2x)^3 )] = ∫du/u^3 +∫du/u^3 = ∫u^(-3)+ ∫u^(-3) = u^(-2)/(-2)+c+u^(-2)/(-2)+c

(2u^(-2))/(-2) = 〖-u〗^(-2)+ c = (1+2x)^(-2)+c = -1/((1+2x)^2 )+c

7. ∫2x(x^2+3)dx

u= x^2+3

du=2xdx

∫u^4 du = 〖 u〗^5/5+c = (〖(x〗^2+3)^5)/5+c

1/5 (x^2+3)^5+c

8. ∫x^3 (1-x^4 )^5 dx

u=1-x^4

du=-4x^3

du/(-4)=x^3

-1/4∫u^5 = -1/4 [u^6/6]+c = -1/4 [((1-x^4 )^6)/6]+c

-((1-x^4 )^6)/24+c

9. ∫√x-1 dx

u=x-1

du=dx

∫u^(1/2) du = u^(3/2)/(3/2)+c = 〖2u〗^(3/2)/3+c

2/3(x-1)^(3/2)+c

10. ∫(2-x)^6 dx

u=2-x

du=-dx

-∫u^6 du = -u^7/7+c

-(2-(x)^7)/7+c

11. ∫dx/(5-3x)

u=5-3x

du=-3dx

du/(-3)=dx

1/(5-3x) dx = -1/3 ∫du/u = -1/3 ln|u|+c

-1/3 ln|5-3x|+c

12. ∫x/(x^2+1) dx

u=x^2+1

du=2x dx

du/2=x dx

1/2∫du/u = 1/2 ln|u|+c

1/2 ln|x^2+1|+c

13. ∫(1+4x)/√(1+x+〖2x〗^2 ) dx

u=1+x+〖2x〗^2

du=1+4x

∫du/u^(1/2) = ∫u^(-1/2) =u^(1/2)/(1/2)+c = 〖2u〗^(1/2)+c = 2(1+x+2x^2 )^(1/2)+c

(2√1+x+2x^2)+ c

14. ∫x(x^2+1)^(3/2) dx

u=x^2+1

du=2x dx

du/u=x dx

1/2∫u^(3/2) du = 1/2 [u^(5/2)/(5/2)]+c = 1/2 [(2u^(5/2))/5]+c = (2u^(5/2))/10+c = 1/5 u^(5/2)+c

1/5

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