# Pattern recognition on integrals questions

1. Jul 3, 2013

### Dan350

Hello users,
I would like to know when do you use pattern recognition over integrals
Someone told me it was that
For example the integral below

I would like to know the procedure to rewrite the numerators as (2x-2) + 3
Where does the 3 come from?
I would really appreciate

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2. Jul 3, 2013

### Ackbach

Pattern recognition is at the heart of mathematics no matter which branch of math you're dealing with. In this case, you can complete the square in the denominator. Rewriting the numerator the way you've mentioned then makes the squared term in the denominator appear in the numerator. Then you can try a change of variables. You get:
$$\int \frac{2x+1}{(x-1)^{2}-3}\,dx = \int \frac{(2x-2)+3}{(x-1)^{2}-3}\,dx =\int \frac{2(x-1)+3}{(x-1)^{2}-3} \, dx.$$
Substitute $u=x-1$ to simplify.

3. Jul 3, 2013

### Dan350

Makes sense,, thank you,, but could you explain me how did you get the number 3 in the numerator??
I know that the integral is in the form f'(x)/f(x),, but what do you do with the constant?
Im a little confused

Last edited: Jul 3, 2013
4. Jul 3, 2013

### Ackbach

Sure. You can't actually change the numerator, can you? You can rewrite it to look a little different, but that's all. The $3$ in the numerator is the number you must add to $-2$ in order to ensure that you end up with the original $+1$. So the thinking goes this way: I want a $-2$ there, but I have a $+1$. How could I get a $-2$? By rewriting $1=-2+3$. So that's where the $3$ comes from.