Partial fraction question.
I have one here that's a cx+d. In my book it says to find c and d I should find 2 x values. I then got A-B = -1/2 2A + 2B = 3
in my book it says to subtract the first from the second so I get 2A - A = A + (2B+B) = 3B = 3--1/2 = 7/2 A+B = 7/2?
The actual question is:
\[\int\limits \frac {x^2 + 2x}{(x-1)(x^2+1)}\]
ummm idk about your book but there shouldn't be a D :/
\[\int\limits \frac{x^2}{(x-1)} + \int\limits \frac{2x}{(x^2+1)} \]
MY book shows D and E, it's just dependent on the exponent values.. When there was x^5 it was A + B + C + Dx + E
but yes for this question what you did is what I have thus far.
I then got it down to \[x^2 + 2x = A(x^2+1) + (Bx+C)(x^2+1)\]
Oh wait I messed up, let me try to fix it.. grr....
what i did so far ... :/
how'd you get that?
doing partial fractions
http://www.wolframalpha.com/input/?i=%E2%88%AB%28x%5E2%2B2x%29%2F%28%28x%E2%88%921%29%28x2%2B1%29%29
it seems like my book keeps them together, but shouldn't I fil out (Bx+C)(x^2+1)?
i re inputted it back into the equation
can you show the steps?
w8 steps for the entire integral or just the partial fraction?
because i can't really help you w/ the integral cuz i'm pretty terrible with them :/
the partial fraction plz.
at first It hought x = 0 to start it off, but I messed up and that doesnt' work...
mhm
hmm i thought there wasnt a square in the problem?
oh wait crap I really messed up LOL let me solve this now...
?
II did (Bx + C) (x^2+1) instead of (x-1)... LOL
if there was a square then the partial fraction would be wrong..
oh w8 I MADE A MISTAKE :0
Iggy I blame you.
okay :( i'll show myself oout :C
w8 no i didnt.....
well @KonradZuse did you get it?
I'm goign to try it again on my own, I understood how to do it and all, but I accidentally messed up(As i said above) so I'm going to try it again because I think it's going to be much easier now.
Okay I think I messed up haha lets step by step this?
It'sd probably because my canvas of choice was ridiculously small boo...
nvm found an error.
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