Odgovor:
#int (-3x + 5) / (x ^ 2-2x + 5) * dx #
# = arctan ((x-1) / 2) -3 / 2ln (x ^ 2-2x + 5) #
Pojasnilo:
#int (-3x + 5) / (x ^ 2-2x + 5) * dx #
=# -int (3x-5) / (x ^ 2-2x + 5) * dx #
=# -int (3x-3-2) / (x ^ 2-2x + 5) * dx #
=# -int (3x-3) / (x ^ 2-2x + 5) * dx #+#int 2 / (x ^ 2-2x + 5) * dx #
=#int 2 / ((x-1) ^ 2 + 4) * dx #-# 3 / 2int (2x-2) / (x ^ 2-2x + 5) #
=#arktan ((x-1) / 2) -3 / 2ln (x ^ 2-2x + 5) #
Odgovor:
# = - 3/2 ln (x ^ 2-2x + 5) + tan ^ -1 ((x-1) / 2) + C #
Pojasnilo:
#int (-3x + 5) / (x ^ 2-2x + 5) dx #
# = int (-3x + 5-2 + 2) / (x ^ 2-2x + 5) dx #
# = int (-3x + 3) / (x ^ 2-2x + 5) + 2 / (x ^ 2-2x + 5) dx #
# = - int (3x-3) / (x ^ 2-2x + 5) dx + int2 / (x ^ 2-2x + 5) dx #
Za:
# -int (3x-3) / (x ^ 2-2x + 5) dx #
Uporabite zamenjavo:
# u = x ^ 2-2x + 5 #
#implies du = 2x-2dx pomeni 3 / 2du = 3x-3dx #
#torej -int (3x-3) / (x ^ 2-2x + 5) dx = -int (3/2) / udu = -3 / 2ln (u) + C #
Obrni zamenjavo:
# -3 / 2ln (x ^ 2-2x + 5) + C #
Zdaj za drugi integral:
# int2 / (x ^ 2-2x + 5) dx #
Vnesite imenovalec v zaključeni kvadratni obliki:
# x ^ 2-2x + 5 = (x-1) ^ 2 - (- 1) ^ 2 + 5 = (x-1) ^ 2 + 4 #
Torej:
# int2 / (x ^ 2-2x + 5) dx = 2intdx / ((x-1) ^ 2 + 4) #
Sedaj nadomestite:
# 2u = (x-1) #
#implies du = 2dx # Torej:
# 2intdx / ((x-1) ^ 2 + 4) = 2int2 / (4u ^ 2 + 4) du = 4 / 4int1 / (u ^ 2 + 1) du #
Za katere bomo prepoznali, bomo preprosto integrirali v inverzno tangento, ki nam daje:
# = tan ^ -1 (u) + C '#
Obrni zamenjavo:
# = tan ^ -1 ((x-1) / 2) + C '#
Zato je "nekaj":
#int (-3x + 5) / (x ^ 2-2x + 5) dx #
# = - int (3x-3) / (x ^ 2-2x + 5) dx + int2 / (x ^ 2-2x + 5) dx #
# = - 3/2 ln (x ^ 2-2x + 5) + tan ^ -1 ((x-1) / 2) + C #
