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..

 

[CellCoree]

Your approach is correct so far. However, you made a small mistake in your second equation. It should be B + 3C = 26 instead of B + C = 26. This will give you a system of equations that can be solved.

A + B = 0
B + 3C = 26
4A + C = 36

Solving this system, you will get A = -4, B = 4, and C = 10. Now you can substitute these values back into the original equation and integrate each term separately.

∫(26x+36)/[(3x-2)(x^2+4)] dx = ∫(-4/(3x-2) + (4x+10)/(x^2+4)) dx

Using the substitution method, you can find the integral of each term. The integral of -4/(3x-2) can be found by using u-substitution. Let u = 3x-2, then du = 3dx and the integral becomes -4/3 ∫1/u du = -4/3 ln|u| + C = -4/3 ln|3x-2| + C.

The integral of (4x+10)/(x^2+4) can be found using partial fractions. Rewrite it as (Ax+B)/(x^2+4) and solve for A and B using the same method you used in the beginning. You will get A = 2 and B = 1. Then the integral becomes ∫(2x+1)/(x^2+4) dx = ∫2/(x^2+4) dx + ∫1/(x^2+4) dx. The first integral can be found using inverse tangent substitution and the second one can be found using u-substitution. The final answer will be -2 tan^-1(x/2) + 1/2 ln|x^2+4| + C.

Therefore, the final answer is -4/3 ln|3x-2| -2 tan^-1(x/2) + 1/2 ln|x^2+4| + C. I hope this helps!