Why does this not work?
#why can't I solve this integral
lilac spindle
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Why does this not work?
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copper fossil
Why does this not work?
you should have, on the 2nd line below the main divide, it should be A(x-2)(x+2), not A(x-2) + A(x+2)
lusty gulch
Also, for denominators with simple roots it's a bit faster to use substitution instead of solving a system.
lilac spindle
hmm , I cant find anywhere where I did A(x-2)(x+2) expect when I was getting the common denominator
copper fossil
hmm , I cant find anywhere where I did A(x-2)(x+2) expect when I was getting the...
this is not correct
lilac spindle
oh let me check again
copper fossil
that is where you have A(x-2) + A(x+2)
instead of A(x-2)(x+2), which is correct
you also made the same mistake with B and C
lilac spindle
I mean I would not be surprised since that part is a little bit obnoxious
Like this ?
But I think I did this already
lusty gulch
If you want to continue with this approach, simplify the polynomials first:
A(x^2 - 4) + B(x^2 + 2x) + C(x^2 - 2x)
lilac spindle
oh yeah this makes no sense
but realistically then would this problem be impossible since its HUGE
lusty gulch
but realistically then would this problem be impossible since its HUGE
As I said, in case of denominators with simple roots substitution is the quicker approach.
But this approach is more universal.
lilac spindle
But what do I do now ? I don't see any way to solve this
lusty gulch
But what do I do now ? I don't see any way to solve this
What are you having trouble with? The usual approach or the substitution approach?
lusty gulch
Well, let's see. We have:
4x + 16 = A(x^2 - 4) + B(x^2 + 2x) + C(x^2 - 2x)
We simplify and group the RHS:
4x + 16 = (A + B + C)x^2 + (2B - 2C)x - 4A
So, what three equations do you get?
lilac spindle
is it
0 = A + B + C
4 = 2B - 2C
16 = -4A
lusty gulch
Yup.
lilac spindle
I mean the previous ones
lusty gulch
Do you want me to show the substitution approach?
lusty gulch
So, we have:
4x + 16 = A(x - 2)(x + 2) + Bx(x + 2) + Cx(x - 2)
We can see that if we substitute one of the roots of the denominator, so x = -2, x = 0 or x = 2, every terms on the RHS except one becomes zero. For example, for x = -2:
4(-2) + 16 = 0 + 0 + C(-2)(-2 - 2)
From here you can easily find C. And you can see that if you substitute another one of the roots, you'll also be able to find A and B without solving a system of equations.
This approach works if the denominator only has simple roots, so nothing like x^2, (x + 3)^3, etc.
There is a modification of that approach for non-simple roots, but it's a bit cumbersome, so in that case it's usually easier to just use the usual undetermined coefficients approach.
lilac spindle
oof yeah I will try to learn that too , it does seem little bit hard , but Im sure its not that bad
and I think I tried to solve this problem in another way
lusty gulch
oof yeah I will try to learn that too , it does seem little bit hard , but Im su...
It's quite useful when there are a lot of simple roots, though.
lilac spindle
yeah will try it for sure
this feels so wrong
is this not a good way to solve this problem in a more easy way
lusty gulch
is this not a good way to solve this problem in a more easy way
This is not correct.
lilac spindle
is it bc I cant have x^2 in denominator ?
lusty gulch
No, you can have it, but only in an irreducible quadratic trinomial, like x^2 + 4 or x^2 + x + 1.
The numerator in that case will be Mx + N, by the way.
lilac spindle
No, you can have it, but only in an irreducible quadratic trinomial, like x^2 + ...
but I have x^2 + 4 in the denominator , is it bc I have x^2 + 4 and something else
lusty gulch
but I have x^2 + 4 in the denominator , is it bc I have x^2 + 4 and something el...
No, you have x^2 - 4, which is obviously not irreducible: x^2 - 4 = (x - 2)(x + 2).
lusty gulch
A quadratic trinomial is irreducible if its discriminant is negative.
lusty gulch
Well, no real solutions.
lilac spindle
so I can do this just if I have something like x^2 - 4
lusty gulch
No. x^2 - 4 = (x - 2)(x + 2), so you can't have it in the denominator.
lilac spindle
okay so I can have only have x^2 + 4 or x^2 + x + 1. in denominator
since I cant factor them more
but if I CAN I MUST factor them
lusty gulch
Yeah.
lilac spindle
okay that's perfect ! I really cannot tell you how much this is helpful . will try to solve more of these types of integrals