Svar:
#int (-3x + 5) / (x ^ 2-2x + 5) * dx #
# = Arctan ((x-1) / 2) -3 / 2ln (x ^ 2-2x + 5) #
Forklaring:
#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) #
=#arctan ((x-1) / 2) -3 / 2ln (x ^ 2-2x + 5) #
Svar:
# = - 3 / 2ln (x ^ 2-2x + 5) + tan ^ -1 ((x-1) / 2) + C #
Forklaring:
#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 #
Til:
# -Int (3x-3) / (x ^ 2-2x + 5) dx #
Bruk substitusjonen:
# U = x ^ 2-2x + 5 #
#implies du = 2x-2dx innebærer 3 / 2du = 3x-3dx #
#therefore -int (3x-3) / (x ^ 2-2x + 5) dx = -int (3/2) / udu = -3/21n (u) + C #
Omvendt erstatning:
# -3 / 2ln (x ^ 2-2x + 5) + C #
Nå for den andre integralen:
# INT2 / (x ^ 2-2x + 5) dx #
Skriv nevneren i fullført kvadratisk form:
# X ^ 2-2x + 5 = (x-1) ^ 2 - (- 1) ^ 2 + 5 = (x-1) ^ 2 + 4 #
Så:
# INT2 / (x ^ 2-2x + 5) dx = 2intdx / ((x-1) ^ 2 + 4) #
Nå erstattet:
# 2u = (x-1) #
#implies du = 2dx # Så:
# 2intdx / ((x-1) ^ 2 + 4) = 2int2 / (4u ^ 2 + 4) du = 4 / 4int1 / (u ^ 2 + 1) du #
Som vi gjenkjenner, vil vi bare integrere til invers tangent som gir oss:
# = Tan ^ -1 (u) + C '#
Omvendt erstatning:
# = Tan ^ -1 ((x-1) / 2) + C '#
Derfor er "noe":
#int (-3x + 5) / (x ^ 2-2x + 5) dx #
# = - int (3x-3) / (x ^ 2-2x + 5) dx + INT2 / (x ^ 2-2x + 5) dx #
# = - 3 / 2ln (x ^ 2-2x + 5) + tan ^ -1 ((x-1) / 2) + C #
