Odgovor:
#int (-3x + 5) / (x ^ 2-2x + 5) * dx #
# = Arctan ((x-1) / 2) -3 / 2ln (x ^ 2-2x + 5) #
Obrazloženje:
#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) #
Odgovor:
# = - 3 / 2ln (x ^ 2-2x + 5) + tan ^ 1 ((x-1) / 2) + C #
Obrazloženje:
#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 (3 x-3), / (x ^ 2-2x + 5) + dx Int2 / (x ^ 2-2x + 5) dx #
Za:
# -Int (3 x-3), / (x ^ 2-2x + 5) dx #
Koristite zamjenu:
# U = x ^ 2-2x + 5 #
#implies du = 2x-2dx podrazumijeva 3 / 2du = 3x-3dx #
#therefore -int (3x-3) / (x ^ 2-2x + 5) dx = -int (3/2) / udu = -3 / 2ln (u) + C #
Obrni zamjenu:
# -3 / 2ln (x ^ 2-2x + 5) + C #
Sada za drugi integral:
# Int2 / (x ^ 2-2x + 5) dx #
Napišite nazivnik u dovršenom kvadratnom obliku:
# X ^ 2-2x + 5 = (x-1) ^ 2 - (- 1) ^ 2 + 5 = (x-1) ^ 2 + 4 #
Tako:
# Int2 / (x ^ 2-2x + 5) dx = 2intdx / ((x-1) ^ 2 + 4) *
Sada zamijenite:
# 2u = (x-1) #
#implies du = 2dx # Tako:
# 2intdx / ((x-1) ^ 2 + 4) = 2int2 / (4U ^ 2 + 4) du = 4 / 4int1 / (z ^ 2 + 1) du #
Što ćemo prepoznati će se jednostavno integrirati u obrnuti tangent koji nam daje:
# = Tan ^ 1 (u) + C '#
Obrni zamjenu:
# = Tan ^ 1 ((x-1) / 2) + C '#
Dakle, "nešto" je:
#int (-3x + 5) / (x ^ 2-2x + 5) dx #
# = - int (3 x-3), / (x ^ 2-2x + 5) + dx Int2 / (x ^ 2-2x + 5) dx #
# = - 3 / 2ln (x ^ 2-2x + 5) + tan ^ 1 ((x-1) / 2) + C #
