How to Integrate Partial Fraction Problems without a Prefix

[Youngster]

Homework Statement

\int\frac{8x^{2}+5x+8}{x^{3}-1}

Homework Equations

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

The Attempt at a Solution

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

7\int\frac{dx}{x-1} + \int\frac{x-1}{x^{2}+x+1}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?

 

[Dick]

Homework Statement

\int\frac{8x^{2}+5x+8}{x^{3}-1}

Homework Equations

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

The Attempt at a Solution

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

7\int\frac{dx}{x-1} + \int\frac{x-1}{x^{2}+x+1}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.

 

[STEMucator]

Complete the square on the other integral. I believe you'll get an arctan in the solution.

 

[Youngster]

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.