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

- inkyvoyd

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

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- chestercat

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- anonymous

Linear by Quadratic...

- inkyvoyd

Eh?

- inkyvoyd

But, how exactly am I supposed to do this one?

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## More answers

- anonymous

If you put ax2 + bx + c = y then it helps or not??

- inkyvoyd

uhh... what about the top - we just substitute using the quadratic formula in that case?

- inkyvoyd

Sorry guys, I'm not even sure where to start with this integral. I can't find it in my book either >.<

- eyust707

bring everything on the bottom to the top... then integration by parts?

- inkyvoyd

uhh - I'm not sure that will work >.<

- eyust707

why not...

- inkyvoyd

All the related integrals had something to do with trig sub and inverse trig functions, but i can't find this case.

- eyust707

ohh

- inkyvoyd

I know what the answer is (from an integral table), and it's logarithmic+trigonometric. I have no idea where to start though

- inkyvoyd

http://integral-table.com/downloads/single-page-integral-table.pdf
#16, btw

- eyust707

ahhh that got me thinkin... let me take another look

- eyust707

well log came from 1/x

- inkyvoyd

Yes.

- eyust707

that looks like a big u sub

- inkyvoyd

Uh, got any ideas where to start htough?

- eyust707

one sec

- inkyvoyd

kay

- eyust707

just by the looks w/o really thinking about it, it looks like they let the bottom = u

- eyust707

then factor out a 2a

- eyust707

1/2a *

- eyust707

wait wtf

- eyust707

lol nvm i thought the second half was the next step

- eyust707

thats a tricky one i need a piece of paper for that one lol

- inkyvoyd

okay lol. Thanks for helping :P

- eyust707

http://www.wolframalpha.com/input/?i=int+x%2F+%28ax2+%2B+bx+%2B+c%29

- eyust707

that way is crazzyyyy tho lol

- inkyvoyd

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

- eyust707

its the same just written different

- experimentX

x = 1/2(2x + b/a) - 1/2 (b/a)
make two fractions ... and use the usual approach

- eyust707

^^^ thats what w.a did too

- inkyvoyd

Okay. any way to reduce it to the form seen in the integral table?

- eyust707

probably not

- inkyvoyd

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.

- experimentX

one is going to be log
another ... complete squares (a+sqrt(b/4a))^2 - (b/4a-c)

- experimentX

this is in standard form ... you should be able to evaluate this.

- inkyvoyd

Wait, I can do it by completing the square as well?

- experimentX

\[ \frac1{2a}\frac{x + b/a}{x + b/a x+c/a} - \frac1{2a}\frac{ b/a}{(x + \sqrt{b/4a})^2 - (b/4a-c/a)}\]

- inkyvoyd

oh wow. this integral - sqrt(tan x) - is a freaking pain...

- experimentX

quite a bit ... assuming c/a > b/4a

- eyust707

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.

- inkyvoyd

Ugh, I give up. I'm tired of this integral lol. I'll look into it more tomorrow. Thanks though, @experimentX and @eyust707

- eyust707

thats when the mathematicians come in handy =P

- eyust707

yw inky.. you taking calc 2 over the summer??

- inkyvoyd

I'm trying to finish calc 3 and linear algebra and diff equations over the summer >.<

- eyust707

crazyyy

- eyust707

that is a lot of material to learn!!

- inkyvoyd

Yea, hopefully I can finish it.

- eyust707

I got faith!! where about are you in diffs?

- inkyvoyd

Haven't even started LOL

- eyust707

haha I see I see

- inkyvoyd

well, I skimmed the variables seperatble and the first order linear ODE I think

- experimentX

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)

- eyust707

yea seps will seem very natural

- eyust707

linear first order with the integrating factor is tricky to understand at first... but easy to do

- eyust707

youll prob see frustrating word problems but dont let them scare you.. they always work out to be quite simple

- inkyvoyd

alright - I have to learn how to integrate these nasty partial fractions first though haha :P

- eyust707

haha just remember, If your integrating with respect to y , all the other variables are constant!

- eyust707

keep repeating that in your head as you run through your derivative..

- eyust707

pretend that all the other letters are just numbers and youll be golden!

- anonymous

didn't you see what i last posted earlier inky?

- inkyvoyd

uhh what did you post?

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