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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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