Question

integral of [(2x^2-1)/((4x-1)(x^2+1))dx]

Answer 1

Is there any particular instructions for solving this?

It looks like Partial Fraction Decomposition is required but I just want to check before we try that.

Answer 2

Yeah, that's what's supposed to be used.

Answer 3

Hmm so the bottom is factored already, that's nice.
Let's see if we can get this broken down correctly.

Answer 4

I've tried it a few times, but the constants are wrong.

Answer 5

So here is our initial setup.
Understand why the B and C are there?
\[\large \frac{2x^2-1}{(4x-1)(x^2+1)} \quad = \quad \frac{A}{4x-1}+\frac{Bx+C}{x^2+1}\]

Answer 6

Yup, that's what I got

Answer 7

Multiplying through by that denominator gives us,\[\large 2x^2-1=A(x^2+1)+(Bx+C)(4x-1)\]

Answer 8

Instead of factoring this all out... let's look for a nice easy X value that might give us some quick information about one of the constants.

Answer 9

if x=1/4?

Answer 10

Ok good! :)
If we let \(\large x=\frac{1}{4}\), That turns the entire Bx+C term to 0.

Answer 11

yup, this is where I keep getting the wrong numbers

Answer 12

\[\large 2\left(\frac{1}{16}\right)-1=A\left(\frac{1}{16}+1\right)\]Ok so far? :o

Answer 13

Yes

Answer 14

A=-14/17?

Answer 15

Hmm yah that sounds right. Is that not what it's suppose to be? :o

Answer 16

-7/34

Answer 17

We made a boo boo somewhere? Hmm ok gimme sec to think :3

Answer 18

oh wait..

Answer 19

It will end up being the constant on top of the 4x-1.
Book says -7/34?

Answer 20

OH

Answer 21

alright, i forgot about the 1/4 from the integral when i was checking it..

Answer 22

hold up for a bit?

Answer 23

k :3

Answer 24

just checking, does B=48/17 and C=3/17?

Answer 25

Pshhhh i dunno, i haven't gotten that far yet XD haha

Answer 26

:O lolk X3

Answer 27

I got C=3/17 and B=18/51.

I should double check my work though... I went through that way too quickly I think.

Answer 28

Alright, I went through it again but B is still wrong (the final constant when you work through it). :/

Answer 29

wait
how did you get B=18/51?

Answer 30

I chose x=0 to solve for C.
Then having solved for A and C, I plugged those into the problem.
From here, I chose x=1 to simplify things down and solved for B.

Answer 31

now i got B=12/17...

Answer 32

ok ok gimme sec :3

Answer 33

I tried it with both
-1=-A+C ---> B=12/17
and
2=A+B ---> B=48/17

T^T

Answer 34

oh you expanded it all out? ew ew :D

Answer 35

XD

Answer 36

It's easier for me to see it that way

Answer 37

\[\large x=1 \qquad \rightarrow \qquad 1=-\frac{14}{17}(2)+\left(B+\frac{3}{17}\right)(3)\]Pshhhh but it plugs in so nicely D":

Answer 38

O_O
It's B=12/17...

Answer 39

Oh so one of your methods worked? :D ooo exciting!

Answer 40

Were the A and C correct? :o

Answer 41

Oh i got 12/17 also, i just didn't calculate correctly the first time.

Answer 42

Aright, when I expanded it, I got:
2=A+B --->(1)
0=-B+4C --->(2)
-1+A-C --->(3)

I got:
A=-14/17
C=3/17
(Both are correct)

When I try to plug A/C into equations (1) or (2) i get different answers.

Am I just doing something wrong?

Answer 43

Yeah, B=12/17 is right., but I'm still curious.

Answer 44

2=A+4B, I don't think you expanded out those right brackets correctly.

Answer 45

You're right...

I'm sorry, thank you.

Answer 46

Heh, easy problem to make a mistake on :D sucks.

Answer 47

I know... -__-

Answer 48

Thank you so much for your help. I really appreciate it. :)