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Homework Statement
Consider the integral:
[tex]\int\frac{2x^3 -4x^2 +8x +7}{(x-1)^2 (x^2 +4x +8)}{\rm{d}}x[/tex]
Homework Equations
The Attempt at a Solution
The degree of the denominator is 4 and the numerator's is 3, hence I thought I would try partial fractions:
[tex]\frac{A}{x-1} +\frac{B}{(x-1)^2} +\frac{C}{x^2 +4x +8} = \frac{2x^3 -4x^2 +8x +7}{(x-1)^2 (x^2 +4x +8)}[/tex]multiplying both sides by the denominator on the right side we would have:
[tex]A(x-1)(x^2 +4x +8) +B(x^2 +4x +8) +C(x-1)^2 = 2x^3 -4x^2 +8x +7\\Ax^3 +(3A +B +C)x^2 +(4A +4B -2C)x -(8A -8B -C) = 2x^3 -4x^2 +8x +7[/tex]
So I should be able to conclude that A = 2, however, the problem is that on one hand I get that 3B = -10 and on the other hand, 10B = 23. Have I made a mistake in the calculations? Is any such rational function divisible [not sure if that's the correct word] into partial fractions?
Is there any other method for tackling such a problem?
Thank you in advance.
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