# Homework Help: Integration Problem

1. Nov 28, 2005

### Briggs

I have a question in my book to evaluate
$$\int_{2}^{3}_\frac{x^2-2x+3}{(x-1)^2}dx$$
I have completed the square on the top line of the fraction so that
$$\frac{x^2-2x+3}{(x-1)^2}$$ = $$\frac{(x-1)^2+2}{(x-1)^2}$$
I have simplified so I end up with $$\int_{2}^{3}1+\frac{2}{(x-1)^2}dx$$ and now I don't know how to integrate this function. I have never encountered an integral with a number other than one on top of the fraction.

Last edited: Nov 28, 2005
2. Nov 28, 2005

### VietDao29

You can use u-substitution in this case.
So let u = x - 1, so du = dx.
The integral becomes:
$$\int \left( 1 + \frac{2}{(x - 1) ^ 2} \right) dx = \int dx + \int \frac{2}{(x - 1) ^ 2} dx = x + \int \frac{2}{u ^ 2} du$$, now after integrating: $$\int \frac{2}{u ^ 2} du$$, just change u back to x. After finding the anti-derivative of the function, just apply the limits to it, and you are done. Can you go from here?

Last edited: Nov 28, 2005
3. Nov 28, 2005

### Briggs

Thanks for the help but the question is from the chapter before the substitution method so I don't think substitution is to be used. Is there another way to do it?

4. Nov 28, 2005

### TD

Perhaps you have used subtitution implicitly. As you said, we can simplify it to:

$$\int {1 + \frac{2}{{\left( {x - 1} \right)^2 }}} dx = \int 1 dx + \int {\frac{2}{{\left( {x - 1} \right)^2 }}dx}$$

Now rewrite to see:

$$\int {\frac{2}{{\left( {x - 1} \right)^2 }}dx} = 2\int {\left( {x - 1} \right)^{ - 2} d\left( {x - 1} \right)}$$

5. Nov 28, 2005

### Briggs

I'm a little confused, would it be..

$$2[\frac{(x-1)^-1}{-1}]$$ with limits 2 and 1? Actually that wouldn't work because of 0^-1, is there any way to get rid of the -1?

Last edited: Nov 28, 2005
6. Nov 28, 2005

### TD

You integrated that second part correctly, but the limits just stay the same as in your original problem, being 2 and 3.

7. Nov 28, 2005

### Briggs

Therefore the answer would be (3+-1)-(2+-2) = 2?

8. Nov 28, 2005

### TD

2 is correct, if you're referring to the entire (initial) integral.

9. Nov 28, 2005

### Briggs

Yes, thank you for your help.

10. Nov 28, 2005

### TD

No problem

11. Nov 28, 2005

### Briggs

Oh by the way, is it correct in saying (X+1)^2 is the same as (X+1)(X-1)

12. Nov 28, 2005

### TD

Certainly not, you can see that by working out both expressions.

Of course, it's easy to see that (x+1)² = (x+1)(x+1) and that's not the same as (x+1)(x-1).
Perhaps you're confused with the identity a²-b² = (a-b)(a+b) with a = x and b = 1.

13. Nov 28, 2005

### Briggs

Yes I was getting mixed up with the difference of two squares because I am trying to solve $$\int_{0}^{1}_\frac{x^2+2x}{(x+1)^2}dx$$
This led me to$$\int_{0}^{1}_\frac{(x-1)^2-1}{(x+1)^2}dx$$ but I have completed the square wrongly and it should be $$(x+1)^2-1$$ which was leading be astray a bit :)

Last edited: Nov 28, 2005
14. Nov 28, 2005

### TD

Ok, but were you able to do it or are you stuck somewhere?

15. Nov 28, 2005

### Briggs

Yep thanks to your help earlier I have completed the question. I got the answer as 0.5

16. Nov 28, 2005

### TD

That is correct