Need integral help ( integral property maybe )

[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.

[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.

[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?

[mathman]

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