- #1
dangish
- 75
- 0
Q: Find the Integral of (26x+36)/[(3x-2)(x^2+4)] dx
Here is my attempt:
First i set it up so that,
(26x+36) / [(3x-2)(x^2+4)] = A/(3x-2) + (Bx+C)/(x^2+4)
Then,
26x + 36 = A(x^2+4) + (Bx+C)(3x-2)
= Ax^2 + 4A + Bx^2+Bx+Cx+C <---(This is where I think I went wrong)
= (A+B)x^2 + (B+C)x + (4A+C)
Then I get the system of equations,
A + B = 0
B + C = 26
4A + C = 36
Which I can't solve.
Once I get A,B,C I can take it from there, any help would be much appreciated :)
Here is my attempt:
First i set it up so that,
(26x+36) / [(3x-2)(x^2+4)] = A/(3x-2) + (Bx+C)/(x^2+4)
Then,
26x + 36 = A(x^2+4) + (Bx+C)(3x-2)
= Ax^2 + 4A + Bx^2+Bx+Cx+C <---(This is where I think I went wrong)
= (A+B)x^2 + (B+C)x + (4A+C)
Then I get the system of equations,
A + B = 0
B + C = 26
4A + C = 36
Which I can't solve.
Once I get A,B,C I can take it from there, any help would be much appreciated :)