Question

\[Solve: \int\limits_{}^{}3\div(2x^2-x-1)dx\]
This is driving me nuts! :(

Answer 1

either try to factor it and use partial fractions, or complete the square then trig sub

Answer 2

I get the answer -2*ln(2x+1)+ln(x-1)+c
Supposedly, the correct answer is ln(x-1)-ln(2x+1)+c

Answer 3

I went the way of partial fractions.

Answer 4

look like you used partial fractions...

Answer 5

...incorrectly

Answer 6

couldn't think of factoring ... looks like factoring.

Answer 7

Haha, trouble is; I can't find where I'm wrong :(

Answer 8

(2x+1)(x-1)

Answer 9

Indeed.

Answer 10

so let's see what you did in the PF part

Answer 11

\[{A\over2x+1}+{B\over x-1}={3\over(2x+1)(x-1)}\implies A(x-1)+B(2x+1)=3\]

Answer 12

so easiest is probably to choose some values for x...
which ones did you choose?

Answer 13

a+2b=0
b-a=3

Answer 14

that's the other way of doing it
which is why you probably messed up
try plugging in x=1 into my post above

Answer 15

either way you get the same answer

Answer 16

The book I'm using only explained it one way. :/

Answer 17

I'm explaining another :)

Answer 18

I do get B=1 and then a=-2

Answer 19

yes you do

Answer 20

Which then - leads to: INT(-2/(2x+1))+INT(1/(x-1))

Answer 21

dx

Answer 22

oh wait you have the right setup the answer is right

Answer 23

But according to both wolf and the book, the other answer is correct, are they the same?
This is crazy! :D

Answer 24

you forgot the u-sub on the first term is all :)

Answer 25

I almost missed it myself for a moment....

Answer 26

Could you elabourate on the u-sub part? I don't follow.

Answer 27

myin will I'm sure

Answer 28

\[\int\limits_{}^{}\frac{1}{2x+1} dx => u=2x+1 => du=2 dx\]

Answer 29

Ah! Thanks so much guys. :)

Answer 30

details, details...
the death of mathematicians everywere

Answer 31

Haha, partial fractal decomposition is so far the most tedious I've done. Takes sooooo long and easy to mess up.

Answer 32

did you understand the method I was describing above?
that way is often easier than solving a system, as you were taught

Answer 33

I think this was is easier than trig sub
Do you know trig sub?

Answer 34

I meant the other way to find the coefficients

Answer 35

I was responding to noliec's comment on partial fractions being most tedious

Answer 36

oh ic

Answer 37

I think so TuringTest! :)
As for myininaya, I don't really know what you mean. :D

Answer 38

But in response to you i'm addicted to solving the system lol

Answer 39

Ok well I'm sure you will figure out in this class
This is like cal 2 right?

Answer 40

Trigonometric substitution? I've done some, not all too much though.
I'm a Swedish student at senior high school.

Answer 41

Hence the trouble converting from "Swedish typestyle" math to English, hehe.

Answer 42

I would do this with http://en.wikipedia.org/wiki/Heaviside_cover-up_method

Answer 43

I will probably go into details of alternative methods in this chapter of the book, we shall see! :)