# Need integral help ( integral property maybe )

1. Oct 4, 2009

### polosportply

What is the $$\int(u)$$-1 where u is a fonction of x , forming a quadratic equation.
As in:

$$\int(u)$$-1 where u = x2+2ax+a2 for example.

Is there a basic property for this... or do I have to play around with the fonction u , in order to integrate?

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I know that $$\int(x)$$ = ln(x) , but this can't be applied here, right? cuz ln(u)' = u-1(u'). Maybe would I need to find a way to eliminate the u' as a result of the integral.

OR

Do I need to change u to = ( x+a)(x+a) and do the integer of that ^-1 , so: $$\int((x+a)^-1)((x+a)^-1)$$

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Does anyone know an integral property for this kind of problem, or know of a way to set me on the right track here?

Thank you.

2. Oct 4, 2009

### mathman

Since you know the integral of x-2 is -x-1, you should be able to do the integral of (x+a)-2. Hint: let y=x+a and get the relationship between dy and dx.

3. Oct 4, 2009

### polosportply

Ok, if I get what you're saying: u$$\equiv(x+a)$$

$$\int(u)$$-2= -u-1= -(x+a)-1

I was onto that before, but then... if I do (-u-1)' = u-2u'

So how come the (u') is not in the equation while doing the integral.
If I integrate something and then derive it, it should come back to the original term.

But apparently not cuz: (u-2)u'$$\neq(u)$$-2

**Is that what you meant mathman, or am I still off?

4. Oct 5, 2009

### mathman

Since u=x+a then u'=1. Therefore u-2u' = u-2