Integrale Inmediatas Cambio De Variables
Enviado por MaribelJimenez • 11 de Marzo de 2012 • 1.313 Palabras (6 Páginas) • 669 Visitas
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
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