Question

Partial fractions integrals! Question in comments.

Answer 1

\[\int\limits_{}^{} \frac{ 2x^2 -x+4 }{ x^3 + 4x }\]

Answer 2

Oooh, looks a bit tricky. Personally, i'd try using the substitution \(u=x^3+4x\) and see if that gets me anywhere... although I'm not saying that it necessarily will :P

Answer 3

They require partial fraction expansion. -_-

Answer 4

Yeah we can expand that out quick..not bad at all for this one

Answer 5

\[\large \frac{2x^2 - x + 4}{x^3 + 4x}\]

Factor out an 'x' from the bottom to begin the process

\[\large \frac{2x^2 - x + 4}{x(x^2 + 4)}\]

Start the partial fraction expansion

\[\large \frac{2x^2 - x + 4}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}\]

Can you continue from there?

Answer 6

I've never seen the Bx + C thing before. How does it work? o_O

Answer 7

If the denominator of your rational expression has an unfactorable quadratic, then you have to account for the possible "size" of the numerator. If the denominator contains a degree-two factor, then the numerator might not be just a number; it might be of degree one. So you would deal with a quadratic factor in the denominator by including a linear expression in the numerator. Hence the reason why you need to have that \(\large Bx + C\) term

Answer 8

So what if we have something like x^3 + 2 as a denominator?

Answer 9

It's safe to say you wont...but it would be the exact same...you would end up with a Bx + C term in the numerator again...it would just be very ugly because you would need to make the denominator a linear term times an irreducible quadratic! Not fun and not done as far as I've ever done

Answer 10

So anytime the denominator ends up with a degree greater than 1, I use Bx + C instead of just a constant?

Answer 11

The worse you may ever have...would be like this:

http://www.purplemath.com/modules/partfrac3.htm
*The first example on that page*

But to answer your question...anytime you have something that can not be factored! You would deal with your linear terms

Answer 12

\[2x^2 -x+4 = A(x^2+4) + (Bx+C)x\]

Answer 13

For example...if your denominator were

\[\large x^2 - 2x - 8\]

That has a degree of 2...but we dont need to deal with linear terms in the numerator because it can be factored!...we can turn that into

\[\large (x - 4)(x + 2)\]

Then its just constants in the numerator

Answer 14

But okay...back to your question

Great! \(\large 2x^2 - x + 4 = A(x^2 + 4) + (Bx + C)(x)\)

Now expand that out fully *distribute what you can*

Answer 15

(A+B)x^2 + Cx + 4A

Answer 16

Good! Equate each side

\[\large 2x^2 - x + 4 = (A+B)x^2 + Cx + 4A\]

So since we can only deal with 'like terms' we have 3 equations and 3 unknowns

\[\large (A+B)x^2 = 2x^2\]
\[\large Cx = -x\]
\[\large 4A = 4\]

What do we have when we solve those?

Answer 17

A = 1, B = 1, C = -1

Answer 18

Exactly!

So now we have what we can integrate:
\[\large \int (\frac{1}{x} + \frac{x - 1}{x^2 + 4})dx\]

Answer 19

\[\int\limits_{}^{}\frac{ 1 }{ x }dx + \int\limits \frac{ x }{ x^2+4 } dx - \int\limits \frac{ 1 }{ x^2 + 4 }dx\]
\[\ln |x| + \frac{ 1 }{ 2 }\ln|x^2 + 4| -\frac{ 1 }{ 2 } \tan ^{-1} \frac{ x }{ 2 } + C\]
Then uhh,

Answer 20

Well I think that's it?

Answer 21

That's what I got \(\large \color \red{\checkmark}\)

Answer 22

I see. Thanks! :D

Answer 23

No problem! Great job!