Hvad er integrationen af (xdx) / sqrt (1-x) ??

Svar:

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

Forklaring:

Lade, # U = sqrt (1-x) #

eller, # U ^ 2 = 1-x #

eller, # X = 1-u ^ 2 #

eller, # Dx = -2udu #

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

Nu, #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 #

Svar:

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

Forklaring:

Integrere efter dele:

#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 #

Svar:

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

Forklaring:

Lade, # 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 #.

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

F.eks, # 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 #.

Hvad er (sqrt (5+) sqrt (3)) / (sqrt (3+) sqrt (3+) sqrt (5)) - (sqrt (5-) sqrt (3)) / (3-) sqrt (5))?

2/7 Vi tager A = (sqrt5 + sqrt3) / (sqrt3 + sqrt3 + sqrt5) - (sqrt5-sqrt3) / (sqrt3 + sqrt3-sqrt5) = (sqrt5 + sqrt3) / (2sqrt3 + sqrt5) - -sqrt3) / (2sqrt3-sqrt5) = (sqrt5 + sqrt3) / (2sqrt3 + sqrt5) - (sqrt5-sqrt3) / (2sqrt3-sqrt5) = ((sqrt5 + sqrt3) (sqrt5-sqrt5) ) (2sqrt3 + sqrt5)) / ((2sqrt3 + sqrt5) (2sqrt3-sqrt5) = ((2sqrt15-5 + 2 * 3-sqrt15) - (2sqrt15 + 5-2 * 3-sqrt15)) / ((2sqrt3) ^ 2- (sqrt5) ^ 2) = (annullere (2sqrt15) -5 + 2 * 3cancel (-sqrt15) - annullere (2sqrt15) -5 + 2 * 3 + annullere (sqrt15)) / (12-5) = ( -10 + 12) / 7 = 2/7 Bemærk, at hvis i betegnelserne er (sqrt3 + sqrt (3 + sqrt5)) og (sqrt3 + sq

Hvad er integrationen af 1 / log (sqrt (1-x))?

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Hvad er integrationen af (dx) / (x.sqrt (x ^ 3 + 4)) ??

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