Thread: Easy partial fractions question

OK, why do you use Bx+c, like where does it come from.

Also when do you use it?

You use partial fractions to integrate rational functions of x (in other words, when you have a polynomial divided by another polynomial). And the best way to think about it is that it's a process that works - for ANY rational function. And I'm sure you're familiar with the process - first divide so you get a proper fraction, then factor the denominator and break apart into partial fractions, then evaluate the parameters, then integrate.

Each step is there to make it so the following steps work. So to answer your question, you use Bx+C when one of the factors of the denominator is an irreducible quadratic - an that doesn't factor into . And you should be able to integrate any function of the form . It might be messy, but you can always do it.

- Hollywood

Thanks, however I still don't get why it works, if I can't factor it, why can I split it into a + (bx+c)?

Firstly when you have a quadratic then you can have at the most a linear function in the numerator that is why Bx + C.
How it works?
Since we cannot factorize the denominator we find the derivative of the denominator which would be a linear function. e.g.,
f(x) = (g(x))/(h(x)) where g(x)is a linear function and h(x) a quadratic function. h’(x) will be a linear function.
Now we will put g(x) = h’(x) + k where k is a constant
Thus we have
f(x) = (g(x))/(h(x))= (h’(x) + k )/(h(x))= (h'(x))/(h(x))+ k/(h(x))
∫▒〖f(x) dx= ∫▒(g(x))/(h(x)) dx〗= ∫▒〖(h’(x) + k )/(h(x)) dx〗= ∫▒〖(h’(x))/(h(x)) dx+ ∫▒k/(h(x)) dx〗
= log⁡〖h(x)〗 + ∫▒k/(h(x)) dx
The last integral can be found by converting the denominator to the form x2 ± a2
????

What does that have to do with Bx + c?

I guess I didn't understand your question. Could you be more specific?

- Hollywood