# Integration of rational functions by partial fractions

1. Sep 10, 2009

### alexas

The problem is:

((x^3)+x)/(x-1)

And i need to break it into partial fractions....

I tried long division and got:

((x^2) +x )

But the book gives me the answer of:

(x^2)+x+2+(2/(x-1))

Any help would be very much appreciated, thanks.

2. Sep 10, 2009

### arildno

Well, IF your long division was right, THEN
x^3+x=(x^2+x)*(x-1)

Is that true?

3. Sep 10, 2009

### arildno

If you are uncertain of the procedure of long division, try to do it as follows:

Since the numerator has highest order 3, and the denominator 1, the highest power of the ratio must be..3-1=2.

We therefore set:

x^3+x=(Ax^2+Bx+C+f(x))*(x-1), where f(x) is some fractional remainder, and A, B, C are constants to be determined.

Multiplying out the right-hand side, and ordering in powers of x, we get:

x^3+x=Ax^3+(B-A)x^2+(C-B)x-C+f(x)*(x-1)

Thus, we must have A=1.

B is therefore also 1, in order to cancel the x^2 term.

C has to be 2, in order for C-B=1

But then, we must have the last equation:
0=-C+f(x)*(x-1),
whereby the remainder f(x)=2/(x-1)

In total then, we get x^2+x+2+2/(x-1) as our answer.