(xdx) / sqrt (1-x)의 통합은 무엇입니까 ??

대답:

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

설명:

방해, # u = sqrt (1-x) #

또는, # u ^ 2 = 1-x #

또는, # x = 1-u ^ 2 #

또는, # dx = -2udu #

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

지금, #int 2u ^ 2 du-int 2du #

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

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

대답:

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

설명:

부품별로 통합:

# 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) #

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

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

(1-x)) + C # (1-x) = -sqrt (1-x)

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

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

대답:

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

설명:

방해, # 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 #.

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

예를 들어, # 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 / 2 + 1) #,

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

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