מהו שילוב של (xdx) / sqrt (1-x) ??

תשובה:

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

הסבר:

תן, # u = sqrt (1-x) #

או, # u ^ 2 = 1-x #

או, # x = 1-u ^ 2 #

או, # dx = -2udu #

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

עכשיו, #int 2u ^ 2 du -int 2du #

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

תשובה:

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

הסבר:

שלב לפי חלקים:

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

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

תשובה:

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

הסבר:

תן, # 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 #, # 1 - int (1-x) ^ (1/2) dx + int (1-x) ^ (- 1/2) dx #.

נזכיר כי, # (x) + dx = F (x) + r rrrr intf (גרף + b) dx = 1 / aF (גרף + b) + K, (a = 0) #

לדוגמה, # 2 / 3x) 3 / 3x (+ 2) 3x). 3/2) + K #.

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