Long division with polynomials

How do I do the integral of x^2/(x-1) doing long division? the answer is supposed to be 1/2 x^2 + x + ln |x-1| Thanks in advance!

...............x + 1
......----------------
x-1 | x^2+0x+ 0
.......x^2 - x
.......----------
...............x + 0
...............x - 1
...............-------
....................1

x^2/(x - 1) = x + 1 + 1/(x - 1)
∫ x^2/(x - 1) dx =
∫ x + 1 = 1/(x - 1) dx =
x^2/2 + x + ln|x - 1| + c

Here is a website for you to reference long division of polynomials.

http://www.khanacademy.org/math/algebra/polynomials/v/dividing-polynomials-1

but for now I'll try to explain....

x-1 goes into x^2 x times. I hope that makes sense.

So multiply x-1 * x and subtract from x^2.

you get a remainder of x. Reapeat with x-1 goes into the remainder x only 1 time. Multiply x-1 *1 and you get a remainder of 1.

So you have 3 parts now, x, 1 and the remainder over the original denomenator, 1/(x-1)

So your new equation is integral (x + 1 + 1/(x-1))

the integral of x is 1/2*x^2 , and the integral of 1 = x, and the integral of 1/(x-1) = 1*ln(x-1)

So the final integration is 1/2x^2 + x + ln(x-1) + constant

I hope this helps.

......... ...x + 1 + 1/(x - 1)
........ ______________
x - 1 |x^2 + 0x + 0
..........x^2 - x
.........---- - - - -
.....................x + 0
.....................x - 1
....................--------
..........................1 (remainder)

Integral[x + 1 + 1/(x - 1) dx] = 1/2 x^2 + x + ln|x - 1| + C