How to Integrate Partial Fraction Problems without a Prefix

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Homework Statement



[itex]\int[/itex][itex]\frac{8x^{2}+5x+8}{x^{3}-1}[/itex]

Homework Equations



Because the denominator can be reduced to (x-1)([itex]x^{2}+x+1[/itex]), I set up the partial fractions to be [itex]\frac{A}{(x-1)}[/itex] + [itex]\frac{Bx+C}{(x^{2}+x+1)}[/itex]

The Attempt at a Solution



I've solved for A, B, and C, and now have the integral set up as such:

7[itex]\int\frac{dx}{x-1}[/itex] + [itex]\int\frac{x-1}{x^{2}+x+1}[/itex]dx

Where A is 7, B is 1, and C is -1

I can integrate the first term simply, but I'm having trouble figuring out how to integrate the second term. The best I can think of is a u substitution, but du turns into 2x+1 dx, which is nothing like x-1 dx. Any suggestions?
 
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Youngster said:

Homework Statement



[itex]\int[/itex][itex]\frac{8x^{2}+5x+8}{x^{3}-1}[/itex]

Homework Equations



Because the denominator can be reduced to (x-1)([itex]x^{2}+x+1[/itex]), I set up the partial fractions to be [itex]\frac{A}{(x-1)}[/itex] + [itex]\frac{Bx+C}{(x^{2}+x+1)}[/itex]

The Attempt at a Solution



I've solved for A, B, and C, and now have the integral set up as such:

7[itex]\int\frac{dx}{x-1}[/itex] + [itex]\int\frac{x-1}{x^{2}+x+1}[/itex]dx

Where A is 7, B is 1, and C is -1

I can integrate the first term simply, but I'm having trouble figuring out how to integrate the second term. The best I can think of is a u substitution, but du turns into 2x+1 dx, which is nothing like x-1 dx. Any suggestions?

Write x-1=(1/2)*(2x+1)-3/2. Now you can easily do the 2x+1 part. The -3/2 part is harder. You'll need to complete the square in the denominator and do a trig substitution.
 
Ah, I see now. It's been a while since I've done that, but it works. I suppose I should do similar exercises to get this in my head.

And yeah, part of the integral turned out to be an inverse tangent one. Thanks a lot.