How Do You Solve the Integral of (26x+36)/[(3x-2)(x^2+4)] dx?

[dangish]

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 :)

 

[rock.freak667]

Take these two

B + C = 26 ...1
4A + C = 36...2

If you subtract equation 1 from 2, you will get an equation with A and B in it, which you can then solve with A + B = 0.

 

[Fragment]

= 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)

Expanding (Bx+C)(3x-2) should give:

3Bx^2 -2Bx +3Cx - 2C

Note that you could have solved for A before expanding this. To do this, notice that (3x-2) can be set to 0 by using x=...?

 

[dangish]

Expanding (Bx+C)(3x-2) should give:

3Bx^2 -2Bx +3Cx - 2C

Haha I was right where I went wrong, that was stupid, thanks..