Foros Ciencias Galilei

1) ∫ t²·ln (t) dt

2) ∫ t·e^-t dt

3) ∫ x²·cos (3x) dx

Gracias, de antemano.

u = t     du = dt

dv = e^-t dt     v = -e^-t

∫u dv = u v - ∫v du


∫ t·e^-t dt = -t·e^-t + ∫e^-t dt = -t·e^-t - e^-t + C   = -e^-t·(t + 1) + C

∫ x²·ln (x) dx = =>=>=>

u = ln x    du =(1/x)dx

dv = x² dx   v = x³/3

=>=>=> = (1/3)·x³·ln x - (1/3)∫x³ (1/x) dx =  (1/3)·x³·ln x - (1/3)∫x² dx = (1/3)·x³·ln x - (1/9)x³ + C

3) ∫ x²·cos (3x) dx =

u = x²     du = 2x dx

dv = cos 3x dx    v = (1/3)·sen 3x

= (1/3)·x²·sen 3x - (2/3)∫x·sen 3x dx =

De nuevo por partes:

u = x    du = dx

dv = sen 3x      v = (-1/3)·cos 3x


= (1/3)·x²·sen 3x - (2/3) [(-x/3)·cos 3x + (1/3)∫cos 3x dx] =

= (1/3)·x²·sen 3x - (2/3) [(-x/3)·cos 3x + (1/9)sen 3x dx] + C

= (1/3)·x²·sen 3x + (2x/9)·cos 3x - (2/27)sen 3x dx + C  =

(1/27) [(9x² - 2)·sen 3x + 6x cos 3x]  +  C