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