Question

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

Answer 1

Linear by Quadratic...

Answer 2

Eh?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

why not...

Answer 10

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

Answer 11

ohh

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

well log came from 1/x

Answer 16

Yes.

Answer 17

that looks like a big u sub

Answer 18

Uh, got any ideas where to start htough?

Answer 19

one sec

Answer 20

kay

Answer 21

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

Answer 22

then factor out a 2a

Answer 23

1/2a *

Answer 24

wait wtf

Answer 25

lol nvm i thought the second half was the next step

Answer 26

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

Answer 27

okay lol. Thanks for helping :P

Answer 28

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

Answer 29

that way is crazzyyyy tho lol

Answer 30

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

Answer 31

its the same just written different

Answer 32

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

Answer 33

^^^ thats what w.a did too

Answer 34

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

Answer 35

probably not

Answer 36

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.

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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.

Answer 44

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

Answer 45

thats when the mathematicians come in handy =P

Answer 46

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

Answer 47

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

Answer 48

crazyyy

Answer 49

that is a lot of material to learn!!

Answer 50

Yea, hopefully I can finish it.

Answer 51

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

Answer 52

Haven't even started LOL

Answer 53

haha I see I see

Answer 54

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

Answer 55

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)

Answer 56

yea seps will seem very natural

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

uhh what did you post?