Kako integrirate int 1 / (x ^ 2 (2x-1)) z uporabo delnih frakcij?

Odgovor:

# 2ln | 2x-1 | -2ln | x | + 1 / x + C #

Pojasnilo:

Moramo najti # A, B, C # tako, da

# 1 / (x ^ 2 (2x-1)) = A / x + B / x ^ 2 + C / (2x-1) #

za vse # x #.

Pomnožite obe strani z # x ^ 2 (2x-1) # dobiti

# 1 = Ax (2x-1) + B (2x-1) + Cx ^ 2 #

# 1 = 2Ax ^ 2-Ax + 2Bx-B + Cx ^ 2 #

# 1 = (2A + C) x ^ 2 + (2B-A) x-B #

Ustrezni koeficienti nam dajo

# {(2A + C = 0), (2B-A = 0), (- B = 1):} #

In tako imamo # A = -2, B = -1, C = 4 #. Če to zamenjamo v začetno enačbo, dobimo

# 1 / (x ^ 2 (2x-1)) = 4 / (2x-1) -2 / x-1 / x ^ 2 #

Zdaj ga vključite v izraz po izrazu

#int 4 / (2x-1) dx-int 2 / x dx-int 1 / x ^ 2 dx #

dobiti

# 2ln | 2x-1 | -2ln | x | + 1 / x + C #

Odgovor:

Odgovor je # = 1 / x-2ln (| x |) + 2ln (| 2x-1 |) + C #

Pojasnilo:

Razgradite v delne frakcije

# 1 / (x ^ 2 (2x-1)) = A / x ^ 2 + B / x + C / (2x-1) #

# = (A (2x-1) + Bx (2x-1) + C (x ^ 2)) / (x ^ 2 (2x-1)) #

Imenovalci so enaki, primerjajte števce

# 1 = A (2x-1) + Bx (2x-1) + C (x ^ 2) #

Let # x = 0 #, #=>#, # 1 = -A #, #=>#, # A = -1 #

Let # x = 1/2 #, #=>#, # 1 = C / 4 #, #=>#, # C = 4 #

Koeficienti # x ^ 2 #

# 0 = 2B + C #

# B = -C / 2 = -4 / 2 = -2

Zato, # 1 / (x ^ 2 (2x-1)) = - 1 / x ^ 2-2 / x + 4 / (2x-1) #

Torej, #int (1dx) / (x ^ 2 (2x-1)) = - int (1dx) / x ^ 2-int (2dx) / x + int (4dx) / (2x-1) #

# = 1 / x-2ln (| x |) + 2ln (| 2x-1 |) + C #