Kaj je integracija (xdx) / sqrt (1-x) ??

Odgovor:

# -2 / 3sqrt (1-x) (2 + x) + C #

Pojasnilo:

Pusti, # u = sqrt (1-x) #

ali, # u ^ 2 = 1-x #

ali, # x = 1-u ^ 2 #

ali, # dx = -2udu #

Zdaj, #int (xdx) / (sqrt (1-x)) = int (1-u ^ 2) (- 2udu) / u = int 2u ^ 2du -int 2du #

Zdaj, #int 2u ^ 2 du -int 2du #

# = (2u ^ 3) / 3 - 2 (u) + C = 2 / 3u (u ^ 2-3) + C = 2 / 3sqrt (1-x) {(1-x) -3} + C = 2 / 3sqrt (1-x) (- 2-x) + C #

# = - 2 / 3sqrt (1-x) (2 + x) + C #

Odgovor:

#int (xdx) / sqrt (1-x) = - (2 (x + 2) sqrt (1-x)) / 3 + C #

Pojasnilo:

Integracija po delih:

#int (xdx) / sqrt (1-x) = int x d (-2sqrt (1-x)) #

#int (xdx) / sqrt (1-x) = -2x sqrt (1-x) + 2 int sqrt (1-x) dx #

#int (xdx) / sqrt (1-x) = -2x sqrt (1-x) - 2 int (1-x) ^ (1/2) d (1-x) #

#int (xdx) / sqrt (1-x) = -2x sqrt (1-x) - 4/3 (1-x) ^ (3/2) + C #

#int (xdx) / sqrt (1-x) = -2x sqrt (1-x) - 4/3 (1-x) sqrt (1-x) + C #

#int (xdx) / sqrt (1-x) = -sqrt (1-x) (2x + 4/3 (1-x)) + C #

#int (xdx) / sqrt (1-x) = -sqrt (1-x) (2 / 3x + 4/3) + C #

#int (xdx) / sqrt (1-x) = - (2 (x + 2) sqrt (1-x)) / 3 + C #

Odgovor:

# -2/3 (2 + x) sqrt (1-x) + C #.

Pojasnilo:

Pusti, # I = intx / sqrt (1-x) dx = -int (-x) / sqrt (1-x) dx #,

# = - int {(1-x) -1} / sqrt (1-x) dx #, # = - int {(1-x) / sqrt (1-x) -1 / sqrt (1-x)} dx #, # = - int {sqrt (1-x) -1 / sqrt (1-x)} dx #, # = - int (1-x) ^ (1/2) dx + int (1-x) ^ (- 1/2) dx #.

Spomnimo se, da

#intf (x) dx = F (x) + C rArr intf (ax + b) dx = 1 / aF (ax + b) + K, (a! = 0) #

Npr. # intx ^ (1/2) dx = 2 / 3x ^ (3/2) + C:.int (2-3x) ^ (1/2) dx = 1 / (- 3) (2-3x) ^ (3/2) + K #.

#:. I = -1 / (- 1) (1-x) ^ (1/2 + 1) / (1/2 + 1) +1 / (- 1) (1-x) ^ (- 1/2 + 1)) / (- 1/2 + 1) #,

# = 2/3 (1-x) ^ (3/2) -2 (1-x) ^ (1/2) #, # = 2/3 (1-x) ^ (1/2) {(1-x) -3} #.

# rArr I = -2 / 3 (2 + x) sqrt (1-x) + C #.