Enkel integreret: int {-3x + 5} / {x ^ 2-2x + 5} dx =?

$Enkel integreret: int {-3x + 5} / {x ^ 2-2x + 5} dx =?$

Svar:

#int (-3x + 5) / (x ^ 2-2x + 5) * dx #

# = Arctan ((x-1) / 2) -3 / 2ln (x ^ 2-2x + 5) #

#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) #

# = - 3 / 2ln (x ^ 2-2x + 5) + tan ^ -1 ((x-1) / 2) + C #

#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 #

Brug substitution:

# U = x ^ 2-2x + 5 #

#implies du = 2x-2dx indebærer 3 / 2du = 3x-3dx #

#therefore -int (3x-3) / (x ^ 2-2x + 5) dx = -int (3/2) / udu = -3/21n (u) + C #

Omvendt udskiftningen:

# -3 / 2ln (x ^ 2-2x + 5) + C #

Nu for den anden integral:

# Int2 / (x ^ 2-2x + 5) dx #

Skriv nævneren i udfyldt 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) #

Nu erstatter:

# 2u = (x-1) #

#implies du = 2dx # Så:

# 2intdx / ((x-1) ^ 2 + 4) = 2int2 / (4u ^ 2 + 4) du = 4 / 4int1 / (u ^ 2 + 1) du #

Det, som vi genkender, vil simpelthen integrere til omvendt tangent, der giver os:

# = Tan ^ -1 (u) + C '#

Omvendt udskiftningen:

# = Tan ^ -1 ((x-1) / 2) + C '#

Derfor er "noget":

#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 #