I have a solution here for the same problem but a different given:
(2x-82)/(x^2+2x-48) dx
In which this solution will help you answer the problem by yourself:
(2x - 82) /(x² + 2x - 48)
first, factor the denominator completely:
(x² + 2x - 48) =
(x² + 8x - 6x - 48) =
x(x + 8) - 6(x + 8) =
(x - 6)(x + 8)
so the function becomes:
(2x - 82) /[(x - 6)(x + 8)]
let's decompose this into partial fractions:
(2x - 82) /[(x - 6)(x + 8)] = A/(x - 6) + B/(x + 8)
(letting [(x - 6)(x + 8)] be the common denominator at the right side too)
(2x - 82) /[(x - 6)(x + 8)] = [A(x + 8) + B(x - 6)] /[(x - 6)(x + 8)]
(equating numerators)
2x - 82 = Ax + 8A + Bx - 6B
2x - 82 = (A + B)x + (8A - 6B)
yielding the system:
A + B = 2
8A - 6B = - 82
A = 2 - B
4A - 3B = - 41
A = 2 - B
4(2 - B) - 3B = - 41
A = 2 - B
8 - 4B - 3B = - 41
A = 2 - B
- 7B = - 41 - 8
A = 2 - B
7B = 49
A = 2 - 7 = - 5
B = 49/7 = 7
hence:
(2x - 82) /[(x - 6)(x + 8)] = A/(x - 6) + B/(x + 8) = - 5/(x - 6) + 7/(x + 8)
thus the answer is:
(2x - 82) /(x² + 2x - 48) = [- 5 /(x - 6)] + [7 /(x + 8)]
