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baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
dw:1358736744745:dw
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.2
\[4x = \frac{ A }{ (x+1) } + \frac{ Bx+C }{ (x^2+1) } + \frac{ Dx+E }{ (x^2+1)^2 }\]
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
thank you that helps alot.. how do you come up with that?
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
and why is it 4x = ......
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
if it is quadratic in denominator in is linear in numerator?
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.2
4x is the numerator. and when it's quadratic it follows the form Ax+B
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
so now we solve all the a,b,c,d,e?
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
how do we solve for all those?
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.2
You get rid of the denominators.
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.2
\[4x = A(x^2+1)^2(x+1) + Bx+C(x+1)(x^2+1) + Dx+E(x+1)\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\large \color{orangered}{\frac{4x}{(x^2+1)^2(x+1)}} = \frac{ A }{ (x+1) } + \frac{ Bx+C }{ (x^2+1) } + \frac{ Dx+E }{ (x^2+1)^2 }\] The left should look like this before you get rid of the denominators :) The rest of that setup looks great though.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
ahh, thank you zepdrix, was a little confused
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
I think i can handle it from here, I just have hard time setting the partial fractions up. any tips?
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.2
Give me a minute to write it all out for you, how in general it works
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
take your time, i'll work on this problem, thanks so much
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
\[4x = A{\left( {{x^2} + 1} \right)^2} + Bx + C\left( {{x^2} + 1} \right)\left( {x + 1} \right) + Dx + E\left( {x + 1} \right)\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
No I guess it was fine, just don't forget the brackets on top.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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...
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
@abb0t did have an extra (x+1) on the first partial..
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Oh i see c:
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.2
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...
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Oh boy... abbot gettin all fancy _ lol
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.2
haters gon hate :P jk :)
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
haha that helps so much... my prof didn't provide us with this info.
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.2
I think it's because you should have learned this in I think algebra or precal. That's why they usually don't go over partial fraction decomp :P
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
:( we had a whole lesson on partial fractions and he seemed to miss "how to split it up"
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Need help finding any of the unknown constants? Or you tearing through this beast just fine? :O
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
haha i had to reboot my comp, but i will slay him now :)
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
or should I expand everything first like you said?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Yah you have to expand them all out :c sucks.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
haha, well, i will give it a shot.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
too bad wolfram wont do it for me
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
I got something ridiculously long
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
I got A=2 I think... Hmm
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Oh you square the square thingy, nevermind 1 sounds correct :)
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
wait i think it's positive 1 now..
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
oh nope.. 1
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
haha, second guessing,
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
I have no idea how to solve the other constants.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\large \begin{align*}4x=&(x^4+2x^2+1)\\&+(Bx^4+Bx^3+Bx^2+Bx\\&+Cx^3+Cx^2+Cx+C)\\&+(Dx^2+Dx+Ex+E)\end{align*}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Did you get something like this? I ran through it a little quickly, so I hope I didn't make any mistakes.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
yep thats what i got.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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,\[\large 0=1+B\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
I should have written \(0x^4\) on the left to stay consistent. Oh well..
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
I don't understand what you mean by 0x^4 and where it should be.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
4x is on the left...
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
ohhhhhh
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
You have NO 4th degree x's on the left. So you equate those with the 4th degree x's on the right.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
:) i get ya now.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
ha ok. I've never heard of equating the like terms before..
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Ya none of my teachers have really explained that step to me very well... So I wouldn't know how to justify it :P
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
dw:1358740342133:dw
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
dw:1358740410506:dw
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
That's kinda funny... I don't think we'll even end up using the 4x to find our unknown constants lol
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
we dont need it to solve for E?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Since there's an Ex and also an E term.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
So either will work.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
oh, 0th degree would be easier... then dw:1358740648463:dw
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
looks good c: I came up with the same values.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
to clarify: A = 1 B = 1 C = 1 D = 2 E = 2
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
\[\large \color{orangered}{\frac{4x}{(x^2+1)^2(x+1)}} = \frac{ 1 }{ (x+1) } + \frac{ x1 }{ (x^2+1) } + \frac{ 2x+2 }{ (x^2+1)^2 }\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
So that's the contents of our new integral? k looks good.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
and now i can integrate it right?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
We're going to end up with a bunch of natural logs and arctangents i think.
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.2
So, what did I miss?
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
dw:1358740971154:dw
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits \frac{1}{x+1} + \color{green}{\frac{x}{x^2+1}\frac{1}{x^2+1}} + \color{cornflowerblue}{\frac{2x}{(x^2+1)^2}+\frac{2}{(x^2+1)^2}}dx\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Understand what I mean about splitting up the fractions?
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
yep, works for me!
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.2
I think it's arctan?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
naw, see how the bottom is being squared? It requires a trig sub and then some fussing with D':
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
ahh, i'll figure it out from here guys, thanks for the help!!!!
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
wish i could give you both medals
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
yay team \c:/
 one year ago
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