Antiderivative of a rational function

[nuuskur]

Homework Statement

Consider the integral:
\int\frac{2x^3 -4x^2 +8x +7}{(x-1)^2 (x^2 +4x +8)}{\rm{d}}x

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:
\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)}multiplying both sides by the denominator on the right side we would have:
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
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.

 

[ehild]

Homework Statement

Consider the integral:
\int\frac{2x^3 -4x^2 +8x +7}{(x-1)^2 (x^2 +4x +8)}{\rm{d}}x

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:
\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)}multiplying both sides by the denominator on the right side we would have:
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
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.

The last partial fraction should contain an additional first-order term in the numerator: \frac{A}{x-1} +\frac{B}{(x-1)^2} +\frac{Cx+D}{x^2 +4x +8} = \frac{2x^3 -4x^2 +8x +7}{(x-1)^2 (x^2 +4x +8)}

 

[nuuskur]

Thank you, ehild, for the correction. Could you please explain why the last partial fraction should also contain the first-order term? I do not doubt your words, I can't fully understand the concept myself.
Everything works out nicely, though.
A = 0, B = 1, C = 2 and D = -1 and the rest is trivial.

 

[mfb]

The denominator is of order 2, and you always need one order less in the numerator (for your (x-1)-type fractions, the A serves that purpose).

As a simple example, you cannot express $\frac{5x+3}{x^2+4x+8}$ with $\frac{C}{x^2+4x+8}$.

 

[ehild]

Could you please explain why the last partial fraction should also contain the first-order term? I do not doubt your words, I can't fully understand the concept myself.

As you have a third-order polynomial in the numerator, it involves 4 equations when comparing its four coefficients with the expression obtained from the partial fractions. So you need 4 unknowns in general. Otherwise you might arrive at contradiction, as you experienced.