Integral by partial fraction?

- baldymcgee6

Integral by partial fraction?

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- baldymcgee6

|dw:1358736744745:dw|

- abb0t

\[4x = \frac{ A }{ (x+1) } + \frac{ Bx+C }{ (x^2+1) } + \frac{ Dx+E }{ (x^2+1)^2 }\]

- baldymcgee6

thank you that helps alot.. how do you come up with that?

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## More answers

- baldymcgee6

and why is it 4x = ......

- baldymcgee6

if it is quadratic in denominator in is linear in numerator?

- abb0t

4x is the numerator.
and when it's quadratic it follows the form Ax+B

- baldymcgee6

so now we solve all the a,b,c,d,e?

- baldymcgee6

how do we solve for all those?

- abb0t

You get rid of the denominators.

- abb0t

\[4x = A(x^2+1)^2(x+1) + Bx+C(x+1)(x^2+1) + Dx+E(x+1)\]

- zepdrix

\[\large \color{orangered}{\frac{4x}{(x^2+1)^2(x+1)}} = \frac{ A }{ (x+1) } + \frac{ Bx+C }{ (x^2+1) } + \frac{ Dx+E }{ (x^2+1)^2 }\]
The left should look like this before you get rid of the denominators :)
The rest of that setup looks great though.

- baldymcgee6

ahh, thank you zepdrix, was a little confused

- baldymcgee6

I think i can handle it from here, I just have hard time setting the partial fractions up. any tips?

- abb0t

Give me a minute to write it all out for you, how in general it works

- baldymcgee6

take your time, i'll work on this problem, thanks so much

- baldymcgee6

\[4x = A{\left( {{x^2} + 1} \right)^2} + Bx + C\left( {{x^2} + 1} \right)\left( {x + 1} \right) + Dx + E\left( {x + 1} \right)\]

- zepdrix

When you multiply through by that denominator you should get,
\[\large \begin{align*} 4x = &A(x^2+1)^2\\&+(Bx+C)(x^2+1)(x+1)\\&+(Dx+E)(x+1) \end{align*}\]
There may have been a small boo boo when abbot multiplied that out, so I just wanted to check.

- zepdrix

No I guess it was fine, just don't forget the brackets on top.

- zepdrix

You can easily solve for A by plugging in x=-1.
That should give you a 0 on the last 2 large terms.
For the rest of them... you have to deal with a nasty squared X term.
So you'll have to expand aaaaaaall of that out :( kinda sucks...

- baldymcgee6

@abb0t did have an extra (x+1) on the first partial..

- zepdrix

Oh i see c:

- abb0t

FOR: \[ax+b \] the term in partial fraction decomposition is: \[\frac{ A }{ ax+b }\] FOR: \[(ax+b)^n\] the term in partial fraction decomposition is: \[\frac{ A_1 }{ ax+b }+\frac{ A_2 }{ (ax+b)^2 }+...+\frac{ A_n }{ (ax+b)^n }\] where n = 1, 2, 3, ...
FOR: \[ax^2+bx+c\] the term in partial fraction decomposition is: \[\frac{ Ax+B }{ ax^2+bx+c }\] and FINALLY, for: \[(ax^2+bx+c)^n\] the partial fration decomposition is: \[\frac{ A_1x+B_1 }{ ax^2+bx+c }+\frac{ A_2x+B_2 }{ (ax^2+bx+c)^2 }+...+\frac{ A_nx+B_n }{ (ax^2+bx+c)^n }\] where same as previously, n = 1, 2, 3...

- zepdrix

Oh boy... abbot gettin all fancy -_- lol

- abb0t

haters gon hate :P
jk :)

- zepdrix

XD

- baldymcgee6

haha that helps so much... my prof didn't provide us with this info.

- abb0t

I think it's because you should have learned this in I think algebra or pre-cal. That's why they usually don't go over partial fraction decomp :P

- baldymcgee6

:( we had a whole lesson on partial fractions and he seemed to miss "how to split it up"

- zepdrix

Need help finding any of the unknown constants? Or you tearing through this beast just fine? :O

- baldymcgee6

haha i had to reboot my comp, but i will slay him now :)

- baldymcgee6

oh man, how do i solve for the constants?? I got A = -1, but i'm not sure what to use for x to solve for the others

- baldymcgee6

or should I expand everything first like you said?

- zepdrix

Yah you have to expand them all out :c sucks.

- zepdrix

I think.

- baldymcgee6

haha, well, i will give it a shot.

- baldymcgee6

too bad wolfram wont do it for me

- baldymcgee6

I got something ridiculously long

- zepdrix

I got A=-2 I think... Hmm

- zepdrix

Oh you square the square thingy, nevermind -1 sounds correct :)

- baldymcgee6

wait i think it's positive 1 now..

- baldymcgee6

oh nope.. -1

- baldymcgee6

haha, second guessing,

- baldymcgee6

I have no idea how to solve the other constants.

- zepdrix

\[\large \begin{align*}4x=&-(x^4+2x^2+1)\\&+(Bx^4+Bx^3+Bx^2+Bx\\&+Cx^3+Cx^2+Cx+C)\\&+(Dx^2+Dx+Ex+E)\end{align*}\]

- zepdrix

Did you get something like this?
I ran through it a little quickly, so I hope I didn't make any mistakes.

- baldymcgee6

yep thats what i got.

- zepdrix

From here, we'll EQUATE LIKE TERMS.
So see how we have \(0x^4\) on the left? we'll set that equal to all the \(x^4\)'s on the right.\[\large 0=-x^4+Bx^4\]Dividing both sides by x gives us,\[\large 0=-1+B\]

- zepdrix

I should have written \(0x^4\) on the left to stay consistent. Oh well..

- baldymcgee6

I don't understand what you mean by 0x^4 and where it should be.

- baldymcgee6

4x is on the left...

- zepdrix

You equate all terms that of the same degree.
There is a 4x on the left. But that is X to the FIRST degree.
We'll equate that with all the of FIRST degree X's on the right.

- baldymcgee6

ohhhhhh

- zepdrix

You have NO 4th degree x's on the left.
So you equate those with the 4th degree x's on the right.

- baldymcgee6

:) i get ya now.

- zepdrix

To be honest the hardest part of the problem is over now :)
Now we just need to be careful,
matching up the orders correctly so we get the right constants.

- baldymcgee6

ha ok. I've never heard of equating the like terms before..

- zepdrix

Ya none of my teachers have really explained that step to me very well...
So I wouldn't know how to justify it :P

- baldymcgee6

|dw:1358740342133:dw|

- baldymcgee6

|dw:1358740410506:dw|

- zepdrix

That's kinda funny... I don't think we'll even end up using the 4x to find our unknown constants lol

- baldymcgee6

we dont need it to solve for E?

- zepdrix

Oh I suppose we could! My bad.
I was thinking of matching up the 0th degree X's to find E.
But we can match up the 1st degree X's also.

- zepdrix

Since there's an Ex and also an E term.

- zepdrix

So either will work.

- baldymcgee6

oh, 0th degree would be easier... then |dw:1358740648463:dw|

- zepdrix

looks good c: I came up with the same values.

- baldymcgee6

to clarify:
A = -1
B = 1
C = -1
D = 2
E = 2

- zepdrix

mhm

- baldymcgee6

\[\large \color{orangered}{\frac{4x}{(x^2+1)^2(x+1)}} = \frac{ -1 }{ (x+1) } + \frac{ x-1 }{ (x^2+1) } + \frac{ 2x+2 }{ (x^2+1)^2 }\]

- zepdrix

So that's the contents of our new integral? k looks good.

- baldymcgee6

and now i can integrate it right?

- zepdrix

Yes, we might have to do a little work to integrate some of the pieces.
Like, we'll have to write the last 2 terms... umm.. we'll need to split up the fractions.

- zepdrix

We're going to end up with a bunch of natural logs and arctangents i think.

- abb0t

So, what did I miss?

- baldymcgee6

|dw:1358740971154:dw|

- zepdrix

\[\int\limits -\frac{1}{x+1} + \color{green}{\frac{x}{x^2+1}-\frac{1}{x^2+1}} + \color{cornflowerblue}{\frac{2x}{(x^2+1)^2}+\frac{2}{(x^2+1)^2}}dx\]

- zepdrix

Understand what I mean about splitting up the fractions?

- baldymcgee6

yep, works for me!

- zepdrix

Yay! So integrate! c:
I'm not actually sure what the last term is going to give us.
I can't think of it off the top of my head lol

- abb0t

I think it's arctan?

- zepdrix

naw, see how the bottom is being squared?
It requires a trig sub and then some fussing with D':

- baldymcgee6

ahh, i'll figure it out from here guys, thanks for the help!!!!

- baldymcgee6

wish i could give you both medals

- zepdrix

yay team \c:/

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