baldymcgee6
  • baldymcgee6
Integral by partial fraction?
Mathematics
jamiebookeater
  • jamiebookeater
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baldymcgee6
  • baldymcgee6
|dw:1358736744745:dw|
abb0t
  • abb0t
\[4x = \frac{ A }{ (x+1) } + \frac{ Bx+C }{ (x^2+1) } + \frac{ Dx+E }{ (x^2+1)^2 }\]
baldymcgee6
  • baldymcgee6
thank you that helps alot.. how do you come up with that?

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baldymcgee6
  • baldymcgee6
and why is it 4x = ......
baldymcgee6
  • baldymcgee6
if it is quadratic in denominator in is linear in numerator?
abb0t
  • abb0t
4x is the numerator. and when it's quadratic it follows the form Ax+B
baldymcgee6
  • baldymcgee6
so now we solve all the a,b,c,d,e?
baldymcgee6
  • baldymcgee6
how do we solve for all those?
abb0t
  • abb0t
You get rid of the denominators.
abb0t
  • abb0t
\[4x = A(x^2+1)^2(x+1) + Bx+C(x+1)(x^2+1) + Dx+E(x+1)\]
zepdrix
  • 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
  • baldymcgee6
ahh, thank you zepdrix, was a little confused
baldymcgee6
  • baldymcgee6
I think i can handle it from here, I just have hard time setting the partial fractions up. any tips?
abb0t
  • abb0t
Give me a minute to write it all out for you, how in general it works
baldymcgee6
  • baldymcgee6
take your time, i'll work on this problem, thanks so much
baldymcgee6
  • 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
  • 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
  • zepdrix
No I guess it was fine, just don't forget the brackets on top.
zepdrix
  • 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
  • baldymcgee6
@abb0t did have an extra (x+1) on the first partial..
zepdrix
  • zepdrix
Oh i see c:
abb0t
  • 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
  • zepdrix
Oh boy... abbot gettin all fancy -_- lol
abb0t
  • abb0t
haters gon hate :P jk :)
zepdrix
  • zepdrix
XD
baldymcgee6
  • baldymcgee6
haha that helps so much... my prof didn't provide us with this info.
abb0t
  • 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
  • baldymcgee6
:( we had a whole lesson on partial fractions and he seemed to miss "how to split it up"
zepdrix
  • zepdrix
Need help finding any of the unknown constants? Or you tearing through this beast just fine? :O
baldymcgee6
  • baldymcgee6
haha i had to reboot my comp, but i will slay him now :)
baldymcgee6
  • 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
  • baldymcgee6
or should I expand everything first like you said?
zepdrix
  • zepdrix
Yah you have to expand them all out :c sucks.
zepdrix
  • zepdrix
I think.
baldymcgee6
  • baldymcgee6
haha, well, i will give it a shot.
baldymcgee6
  • baldymcgee6
too bad wolfram wont do it for me
baldymcgee6
  • baldymcgee6
I got something ridiculously long
zepdrix
  • zepdrix
I got A=-2 I think... Hmm
zepdrix
  • zepdrix
Oh you square the square thingy, nevermind -1 sounds correct :)
baldymcgee6
  • baldymcgee6
wait i think it's positive 1 now..
baldymcgee6
  • baldymcgee6
oh nope.. -1
baldymcgee6
  • baldymcgee6
haha, second guessing,
baldymcgee6
  • baldymcgee6
I have no idea how to solve the other constants.
zepdrix
  • 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
  • zepdrix
Did you get something like this? I ran through it a little quickly, so I hope I didn't make any mistakes.
baldymcgee6
  • baldymcgee6
yep thats what i got.
zepdrix
  • 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
  • zepdrix
I should have written \(0x^4\) on the left to stay consistent. Oh well..
baldymcgee6
  • baldymcgee6
I don't understand what you mean by 0x^4 and where it should be.
baldymcgee6
  • baldymcgee6
4x is on the left...
zepdrix
  • 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
  • baldymcgee6
ohhhhhh
zepdrix
  • 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
  • baldymcgee6
:) i get ya now.
zepdrix
  • 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
  • baldymcgee6
ha ok. I've never heard of equating the like terms before..
zepdrix
  • 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
  • baldymcgee6
|dw:1358740342133:dw|
baldymcgee6
  • baldymcgee6
|dw:1358740410506:dw|
zepdrix
  • zepdrix
That's kinda funny... I don't think we'll even end up using the 4x to find our unknown constants lol
baldymcgee6
  • baldymcgee6
we dont need it to solve for E?
zepdrix
  • 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
  • zepdrix
Since there's an Ex and also an E term.
zepdrix
  • zepdrix
So either will work.
baldymcgee6
  • baldymcgee6
oh, 0th degree would be easier... then |dw:1358740648463:dw|
zepdrix
  • zepdrix
looks good c: I came up with the same values.
baldymcgee6
  • baldymcgee6
to clarify: A = -1 B = 1 C = -1 D = 2 E = 2
zepdrix
  • zepdrix
mhm
baldymcgee6
  • 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
  • zepdrix
So that's the contents of our new integral? k looks good.
baldymcgee6
  • baldymcgee6
and now i can integrate it right?
zepdrix
  • 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
  • zepdrix
We're going to end up with a bunch of natural logs and arctangents i think.
abb0t
  • abb0t
So, what did I miss?
baldymcgee6
  • baldymcgee6
|dw:1358740971154:dw|
zepdrix
  • 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
  • zepdrix
Understand what I mean about splitting up the fractions?
baldymcgee6
  • baldymcgee6
yep, works for me!
zepdrix
  • 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
  • abb0t
I think it's arctan?
zepdrix
  • zepdrix
naw, see how the bottom is being squared? It requires a trig sub and then some fussing with D':
baldymcgee6
  • baldymcgee6
ahh, i'll figure it out from here guys, thanks for the help!!!!
baldymcgee6
  • baldymcgee6
wish i could give you both medals
zepdrix
  • zepdrix
yay team \c:/

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