Question

Hello.

\[\LARGE \int \frac{1}{(x^2+x-2)}dx\]

now..
\[\LARGE \int \frac{1}{(x+2)(x-1 )}dx\]
if I have to split them like:
\[\LARGE \frac{1}{(x+2)(x-1 )}=\frac{A}{(x+2)}+\frac{B}{(x-1 )}\]
I've come up with...
1=x(A+B)+2B-A
but I don't know what to do any further...

is there anything wrong...
I know the answer should be:
\[\LARGE \frac13\ln\left|\frac{x-1}{x+2}\right|\]
can someone please help me integrate this? :)

Answer 1

I will let someone pick up the partial fractions way, but I think that it can be done completing the square first: \[\int \frac{1}{(x + 1/2)^2 - \frac{9}{4}}\]Then u = x + 1/2, that becomes an hyperbolic inverse, I think, arctanh.

Answer 2

let me see what I can do... :)

Answer 3

if you ended up with:
1=x(A+B)+2B-A

then this implies:
A+B=0
2B-A=1

Answer 4

really? ...hmm, ok , as long as I need some practice with these A+B , I'll choose this way ... I think the rest I can do, I just didn't know how to make a system thanks @asnaseer thanks @bmp . :)

Answer 5

the reason A+B=0 is because you have no 'x' on the left-hand-side

Answer 6

so effectively you have:

0*x + 1 = (A+B)*x + (2B-A)

Answer 7

yeah I know ;) ... but I had never something similar, and that's the reason why I got stuck :) thanks again.

Answer 8

yw

Answer 9

Why don't you use heavy side cover up ( http://en.wikipedia.org/wiki/Heaviside_cover-up_method) ? It's fast!

Answer 10

I'm just a beginner at integrals, I'm cool with this lol haha.. ;)

Answer 11

interesting FFM - I never heard of this method - thx :)

Answer 12

yw @asnaseer :) and @Kreshnik it has nothing to do with integrals. Partial fraction decomposition is actually algebra-pre-calculus :)

Answer 13

well then I'm a beginner at that too !! ... ^^