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Integral by partial fraction?

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

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Other answers:

and why is it 4x = ......
if it is quadratic in denominator in is linear in numerator?
4x is the numerator. and when it's quadratic it follows the form Ax+B
so now we solve all the a,b,c,d,e?
how do we solve for all those?
You get rid of the denominators.
\[4x = A(x^2+1)^2(x+1) + Bx+C(x+1)(x^2+1) + Dx+E(x+1)\]
\[\large \color{orangered}{\frac{4x}{(x^2+1)^2(x+1)}} = \frac{ A }{ (x+1) } + \frac{ Bx+C }{ (x^2+1) } + \frac{ Dx+E }{ (x^2+1)^2 }\] The left should look like this before you get rid of the denominators :) The rest of that setup looks great though.
ahh, thank you zepdrix, was a little confused
I think i can handle it from here, I just have hard time setting the partial fractions up. any tips?
Give me a minute to write it all out for you, how in general it works
take your time, i'll work on this problem, thanks so much
\[4x = A{\left( {{x^2} + 1} \right)^2} + Bx + C\left( {{x^2} + 1} \right)\left( {x + 1} \right) + Dx + E\left( {x + 1} \right)\]
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.
No I guess it was fine, just don't forget the brackets on top.
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...
@abb0t did have an extra (x+1) on the first partial..
Oh i see c:
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...
Oh boy... abbot gettin all fancy -_- lol
haters gon hate :P jk :)
XD
haha that helps so much... my prof didn't provide us with this info.
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
:( we had a whole lesson on partial fractions and he seemed to miss "how to split it up"
Need help finding any of the unknown constants? Or you tearing through this beast just fine? :O
haha i had to reboot my comp, but i will slay him now :)
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
or should I expand everything first like you said?
Yah you have to expand them all out :c sucks.
I think.
haha, well, i will give it a shot.
too bad wolfram wont do it for me
I got something ridiculously long
I got A=-2 I think... Hmm
Oh you square the square thingy, nevermind -1 sounds correct :)
wait i think it's positive 1 now..
oh nope.. -1
haha, second guessing,
I have no idea how to solve the other constants.
\[\large \begin{align*}4x=&-(x^4+2x^2+1)\\&+(Bx^4+Bx^3+Bx^2+Bx\\&+Cx^3+Cx^2+Cx+C)\\&+(Dx^2+Dx+Ex+E)\end{align*}\]
Did you get something like this? I ran through it a little quickly, so I hope I didn't make any mistakes.
yep thats what i got.
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,\[\large 0=-1+B\]
I should have written \(0x^4\) on the left to stay consistent. Oh well..
I don't understand what you mean by 0x^4 and where it should be.
4x is on the left...
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.
ohhhhhh
You have NO 4th degree x's on the left. So you equate those with the 4th degree x's on the right.
:) i get ya now.
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.
ha ok. I've never heard of equating the like terms before..
Ya none of my teachers have really explained that step to me very well... So I wouldn't know how to justify it :P
|dw:1358740342133:dw|
|dw:1358740410506:dw|
That's kinda funny... I don't think we'll even end up using the 4x to find our unknown constants lol
we dont need it to solve for E?
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.
Since there's an Ex and also an E term.
So either will work.
oh, 0th degree would be easier... then |dw:1358740648463:dw|
looks good c: I came up with the same values.
to clarify: A = -1 B = 1 C = -1 D = 2 E = 2
mhm
\[\large \color{orangered}{\frac{4x}{(x^2+1)^2(x+1)}} = \frac{ -1 }{ (x+1) } + \frac{ x-1 }{ (x^2+1) } + \frac{ 2x+2 }{ (x^2+1)^2 }\]
So that's the contents of our new integral? k looks good.
and now i can integrate it right?
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.
We're going to end up with a bunch of natural logs and arctangents i think.
So, what did I miss?
|dw:1358740971154:dw|
\[\int\limits -\frac{1}{x+1} + \color{green}{\frac{x}{x^2+1}-\frac{1}{x^2+1}} + \color{cornflowerblue}{\frac{2x}{(x^2+1)^2}+\frac{2}{(x^2+1)^2}}dx\]
Understand what I mean about splitting up the fractions?
yep, works for me!
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
I think it's arctan?
naw, see how the bottom is being squared? It requires a trig sub and then some fussing with D':
ahh, i'll figure it out from here guys, thanks for the help!!!!
wish i could give you both medals
yay team \c:/

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