# How did the book do this integration

1. Oct 18, 2004

### kdinser

I'm sure it's pretty simple, but I'm just not seeing it.

Integrate (x^2/(x-1)) dx

I did it with a u substitution, letting u = x-1 and then x = u+1

which ultimately leads me to integrate (u^2/u) + (2u/u) + (1/u)

After canceling, integrating, and substituting I'm left with
(x^2/2) + x + ln(abs)(x-1) + C (I assume I'm alright rolling the -3/2 that I had left over into C to make it match the book answer?)

The book does it like this and I'm not sure what's going on
Integrate (x^2/(x-1)) dx = Integrate (x+1) dx + Integrate (1/(x-1)) dx

I think they are splitting it up somehow, hence the 1/x-1, but I'm not sure how they got to x+1

which yields the same answer I had.

2. Oct 18, 2004

### Fredrik

Staff Emeritus
$$x^2=x^2-1+1=(x+1)(x-1)+1$$

3. Oct 18, 2004

### Tide

All they are doing is rewriting the integrand this way:

$$\frac {x^2}{x-1} = \frac {x^2 -1 + 1}{x-1} = \frac {x^2-1}{x-1}+\frac {1}{x-1} = x+1 + \frac {1}{x-1}$$

4. Oct 18, 2004

### kdinser

Bah, I must have been way off my game this morning, thanks.