∫▒(xe^2x)/(1+2x)^2
Hola
ʃ x e^(2x) dx/(1+ 2 x)^2
y = 2 x
x = (1/2) y
dy = 2 dx
dx = (1/2) dy
ʃ x e^(2x) dx/(1+ 2 x)^2 =(1/4) ʃ y e^y dy/(1+ y)^2
Integración por partes
u = - y e^y
v = 1/(1+y)
dv = -dy/(1 + y)^2
du = d(-y e^y) = (-1)(y e^y + e^y) dy = (-1)(1 + y) e^y dy
Queda
ʃ x e^(2x) dx/(1+ 2 x)^2 =(1/4) ʃ u dv
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) u v - (1/4) ʃ v du
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) (-y e^y) (1/(1 + y)) +
+ (1/4) ʃ (1/(1 + y)) (1 + y) e^y dy
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) (e^y) (-y/(1 + y)) +
+ (1/4) ʃ e^y dy
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) (e^y) (-y/(1 + y)) + (1/4) e^y + C
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) (e^y) (-y + 1 + y)/(1 + y)) + C
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) (e^y) (1)/(1 + y)) + C
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) (e^(2x))/(1 + 2x)) + C
***************************************************************
Saludos
vuelvo
No voy a hacer tu tarea mientras te rascas.
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Answers & Comments
Hola
ʃ x e^(2x) dx/(1+ 2 x)^2
y = 2 x
x = (1/2) y
dy = 2 dx
dx = (1/2) dy
ʃ x e^(2x) dx/(1+ 2 x)^2 =(1/4) ʃ y e^y dy/(1+ y)^2
Integración por partes
u = - y e^y
v = 1/(1+y)
dv = -dy/(1 + y)^2
du = d(-y e^y) = (-1)(y e^y + e^y) dy = (-1)(1 + y) e^y dy
Queda
ʃ x e^(2x) dx/(1+ 2 x)^2 =(1/4) ʃ u dv
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) u v - (1/4) ʃ v du
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) (-y e^y) (1/(1 + y)) +
+ (1/4) ʃ (1/(1 + y)) (1 + y) e^y dy
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) (e^y) (-y/(1 + y)) +
+ (1/4) ʃ e^y dy
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) (e^y) (-y/(1 + y)) + (1/4) e^y + C
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) (e^y) (-y + 1 + y)/(1 + y)) + C
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) (e^y) (1)/(1 + y)) + C
ʃ x e^(2x) dx/(1+ 2 x)^2 = (1/4) (e^(2x))/(1 + 2x)) + C
***************************************************************
Saludos
vuelvo
No voy a hacer tu tarea mientras te rascas.