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Integrate \(\Huge \frac{x}{ax^2+bx+c}\)

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Linear by Quadratic...
But, how exactly am I supposed to do this one?

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Other answers:

If you put ax2 + bx + c = y then it helps or not??
uhh... what about the top - we just substitute using the quadratic formula in that case?
Sorry guys, I'm not even sure where to start with this integral. I can't find it in my book either >.<
bring everything on the bottom to the top... then integration by parts?
uhh - I'm not sure that will work >.<
why not...
All the related integrals had something to do with trig sub and inverse trig functions, but i can't find this case.
I know what the answer is (from an integral table), and it's logarithmic+trigonometric. I have no idea where to start though #16, btw
ahhh that got me thinkin... let me take another look
well log came from 1/x
that looks like a big u sub
Uh, got any ideas where to start htough?
one sec
just by the looks w/o really thinking about it, it looks like they let the bottom = u
then factor out a 2a
1/2a *
wait wtf
lol nvm i thought the second half was the next step
thats a tricky one i need a piece of paper for that one lol
okay lol. Thanks for helping :P
that way is crazzyyyy tho lol
lol - but the answer is different from the integral table :S the result is too complicated lol. I'm using some constants for the integral that will make it even more nasty x.x
its the same just written different
x = 1/2(2x + b/a) - 1/2 (b/a) make two fractions ... and use the usual approach
^^^ thats what w.a did too
Okay. any way to reduce it to the form seen in the integral table?
probably not
once i substitute my values in, it is still an intermediate result - I still have to do two more substitutions, and I'm not sure I can do the algebra without going insane.
one is going to be log another ... complete squares (a+sqrt(b/4a))^2 - (b/4a-c)
this is in standard form ... you should be able to evaluate this.
Wait, I can do it by completing the square as well?
\[ \frac1{2a}\frac{x + b/a}{x + b/a x+c/a} - \frac1{2a}\frac{ b/a}{(x + \sqrt{b/4a})^2 - (b/4a-c/a)}\]
oh wow. this integral - sqrt(tan x) - is a freaking pain...
quite a bit ... assuming c/a > b/4a
yea just about all realistic functions are either extremely tedious or impossible to solve by hand. When it comes to practical application, we usually just have computers solve or approximate the solution.
Ugh, I give up. I'm tired of this integral lol. I'll look into it more tomorrow. Thanks though, @experimentX and @eyust707
thats when the mathematicians come in handy =P
yw inky.. you taking calc 2 over the summer??
I'm trying to finish calc 3 and linear algebra and diff equations over the summer >.<
that is a lot of material to learn!!
Yea, hopefully I can finish it.
I got faith!! where about are you in diffs?
Haven't even started LOL
haha I see I see
well, I skimmed the variables seperatble and the first order linear ODE I think
It's just a substitution ... reduces to the form \[ \int \frac1 {x^2 + a^2}dx\] or, \[ \int \frac 1{x^2 - a^2}dx\] assuming first, wolf calculates 2/a arctan(x/a)
yea seps will seem very natural
linear first order with the integrating factor is tricky to understand at first... but easy to do
youll prob see frustrating word problems but dont let them scare you.. they always work out to be quite simple
alright - I have to learn how to integrate these nasty partial fractions first though haha :P
haha just remember, If your integrating with respect to y , all the other variables are constant!
keep repeating that in your head as you run through your derivative..
pretend that all the other letters are just numbers and youll be golden!
didn't you see what i last posted earlier inky?
uhh what did you post?

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