A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1358736744745:dw

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

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0thank you that helps alot.. how do you come up with that?

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0and why is it 4x = ......

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0if it is quadratic in denominator in is linear in numerator?

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.24x is the numerator. and when it's quadratic it follows the form Ax+B

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0so now we solve all the a,b,c,d,e?

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0how do we solve for all those?

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2You get rid of the denominators.

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

zepdrix
 2 years ago
Best 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.

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0ahh, thank you zepdrix, was a little confused

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

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

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0take your time, i'll work on this problem, thanks so much

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0When 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
 2 years ago
Best ResponseYou've already chosen the best response.0No I guess it was fine, just don't forget the brackets on top.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0You 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
 2 years ago
Best ResponseYou've already chosen the best response.0@abb0t did have an extra (x+1) on the first partial..

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2FOR: \[ax+b \] 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
 2 years ago
Best ResponseYou've already chosen the best response.0Oh boy... abbot gettin all fancy _ lol

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0haha that helps so much... my prof didn't provide us with this info.

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.2I 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

baldymcgee6
 2 years ago
Best 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"

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

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0haha i had to reboot my comp, but i will slay him now :)

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0oh 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
 2 years ago
Best ResponseYou've already chosen the best response.0or should I expand everything first like you said?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Yah you have to expand them all out :c sucks.

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0haha, well, i will give it a shot.

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0too bad wolfram wont do it for me

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0I got something ridiculously long

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0I got A=2 I think... Hmm

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Oh you square the square thingy, nevermind 1 sounds correct :)

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0wait i think it's positive 1 now..

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0haha, second guessing,

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0I have no idea how to solve the other constants.

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

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

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0yep thats what i got.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0From 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
 2 years ago
Best ResponseYou've already chosen the best response.0I should have written \(0x^4\) on the left to stay consistent. Oh well..

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

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.04x is on the left...

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0You 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.

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0To 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
 2 years ago
Best ResponseYou've already chosen the best response.0ha ok. I've never heard of equating the like terms before..

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

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1358740342133:dw

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1358740410506:dw

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

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0we dont need it to solve for E?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Oh 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
 2 years ago
Best ResponseYou've already chosen the best response.0Since there's an Ex and also an E term.

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0oh, 0th degree would be easier... then dw:1358740648463:dw

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0looks good c: I came up with the same values.

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0to clarify: A = 1 B = 1 C = 1 D = 2 E = 2

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0So that's the contents of our new integral? k looks good.

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0and now i can integrate it right?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 2 years ago
Best ResponseYou've already chosen the best response.0We're going to end up with a bunch of natural logs and arctangents i think.

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1358740971154:dw

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Understand what I mean about splitting up the fractions?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Yay! 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

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

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0ahh, i'll figure it out from here guys, thanks for the help!!!!

baldymcgee6
 2 years ago
Best ResponseYou've already chosen the best response.0wish i could give you both medals
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.