Int2 / (2x ^ 2 + 2x) dx?

Odgovor:

#ln (abs (x / (x + 1))) + C #

Pojasnilo:

Najprej izračunamo 2:

# int1 / (x ^ 2 + x) dx #

Potem faktorizirajte imenovalec:

# int1 / (x (x + 1)) dx #

Razdeliti moramo na delne dele:

# 1 = A (x + 1) + Bx #

Uporaba # x = 0 # nam daje:

# A = 1 #

Potem z uporabo # x = -1 # nam daje:

# 1 = -B #

S tem dobimo:

# int1 / x-1 / (x + 1) dx #

# int1 / xdx-int / (x + 1) dx #

#ln (abs (x)) - ln (abs (x + 1 _) + C #

#ln (abs (x / (x + 1))) + C #