Question

Integration by Partial Fractions!

Answer 1

\[\int\limits \frac{ 2x-1 }{ (4x-1)(x^2+1) } dx\]

Answer 2

Ok this one is a lil sneaky!
See how we have a `quadratic factor` (2nd degree on x) that can't be broken down any further?

That means the fraction it forms will have a `linear` (1st degree on x) in the numerator, not just a constant.

Answer 3

Got it!

Answer 4

\[\Large \frac{2x-1}{(4x-1)(x^2+1)}\quad=\quad \frac{A}{(4x-1)}+\frac{Bx+C}{x^2+1}\]
Make sense, kinda? :o
See how that numerator is linear?

Answer 5

So far, so good!

Answer 6

It'll be relatively the same process at before.
Ummm since we have the sum of squares, we might wanna take the `other` approach to this problem ~ Expanding everything out.

Answer 7

Got it!

Answer 8

\[\Large 2x-1 \quad=\quad A(x^2+1)+B(4x-1)\]

Answer 9

Got it!

Answer 10

What happened with the C?

Answer 11

Oh goodness I'm so silly lolol :)

Answer 12

Yes yes good catch :3\[\Large 2x-1 \quad=\quad A(x^2+1)+(Bx+C)(4x-1)\]

Answer 13

So what do you get when you expand all that out? XD

Answer 14

I would use \[i\] and \[\frac{1}{4}\]

Answer 15

breaking it up into complex conjugates?
ehhhh :3

Answer 16

When I expand it, I get Ax^2 + 4Bx + 4Cx + A -B - C

Answer 17

multiplying it out is fine too...just thought complex numbers would be fun ;)

Answer 18

XD

Answer 19

zepdrix

When I expand it, I get Ax^2 + 4Bx + 4Cx + A -B - C

Answer 20

\[\large 2x-1 \quad=\quad \color{orangered}{A(x^2+1)}+\color{royalblue}{(Bx+C)(4x-1)}\]Expanding should give us,\[\large 2x-1 \quad=\quad \color{orangered}{Ax^2+A}+\color{royalblue}{4Bx^2-Bx+4Cx-C}\]

Answer 21

Expanding the blue part is a little tricky, maybe take another look at it :o

Answer 22

Oh, I see!

Answer 23

Personally, I would recommend writing it like this:\[\large 0x^2+2x-1 \quad=\quad Ax^2+A+4Bx^2-Bx+4Cx-C\]with a 0x^2 on the left, so you can see how the x^2's will equate with that side.

Answer 24

Got it!

Answer 25

We have x^2's, x's, and constants's.
So we'll have a system of 3 equations.
Can you set them up properly? :D

Answer 26

Tricky question...

Answer 27

You equate like powers.
So lemme show ya how the x^2's will work out.
Equating the left side's x^2's with the right side's gives us:\[\Large 0x^2 \quad=\quad Ax^2+4Bx^2\]
Dividing both sides by x^2 gives us,\[\Large 0\quad=\quad A+4B\]

Answer 28

So there is the first of our 3 equations.
Think you can do it with the x's?

Answer 29

Oh, okay! Let me try it!

Answer 30

0 = B + 4C
0 = A - C

Answer 31

Woops you didn't use the values from the left side :(
So let's look at the x's a moment, then we'll do the constants after that.

\[\Large 2x \quad=\quad -Bx+4x\]Understand how I set that up?

Answer 32

Woops typo,\[\Large 2x \quad=\quad -Bx+4Cx\]

Answer 33

-1=A-C

Answer 34

Ya looks good :)
As we did in the x^2's, we want to ignore/divide out the actual x terms.
So the second equation gives us 2=-B+4

Answer 35

Yeah, I got it!

Answer 36

Grr I missed the C on my 4C again lol. Anyway,

Our system is:\[\Large 0=A+4B\]\[\Large 2=-B+4C\]\[\Large -1=A-C\]

Answer 37

So try to solve for A, B and C from here.

Answer 38

A = -1/4
B = 1/4
C = 3/4

Answer 39

Bah one sec :3 lemme throw it into my handy calculator real quick to check.

Answer 40

Okay!

Answer 41

Hmm I'm coming up with:
A=-8/17
B=2/17
C=9/17

Maybe I punched it in wrong, lemme try it on paper a sec >.<

Answer 42

A+B should equal zero.

Answer 43

You mean A+4B? :o

Answer 44

I simplified it!
0 = A+B
1/2 =-B+C
-1=A-C

Answer 45

Then I thrown it into my calculator.

Answer 46

You made a mistake simplifying I'm afraid :(

\[\Large 0=A+4B\]Dividing by 4 gives us,\[\Large 0=\frac{1}{4}A+B\]

Answer 47

You don't have to be afraid! I just over-looked the problem! But I got it!

Answer 48

lol :3

Answer 49

Got it!

Answer 50

A=-8/17
B=2/17
C=9/17

Answer 51

?

Answer 52

??

Answer 53

?

Answer 54

Is someone still there?

Answer 55

Oh sorry was making some popcorn XD lol

Answer 56

So you tried it again and those were the numbers you came up with?
Or you just copied my answers? XD lol

Answer 57

I tried it again anod those are numbers I came up with!

Answer 58

and*

Answer 59

k cool c:

Answer 60

\[\large \int\limits\limits \frac{ 2x-1 }{ (4x-1)(x^2+1) } dx \quad=\quad \int\limits \frac{A}{4x-1}+\frac{Bx+C}{x^2+1}dx\]

Answer 61

I guess it would make sense to factor a 1/17 out of everything right?
make's it quite a bit easier to read.\[\Large \frac{1}{17}\int\limits \frac{-8}{4x-1}+\frac{2x+9}{x^2+1}dx\]Understand what I did there?

Answer 62

Yeah! That's what I did, too!

Answer 63

cool

Answer 64

So now from here, before we integrate, you'll want to break it down just a tad further.
We can do this to the second term:\[\Large \frac{1}{17}\int\limits\limits \frac{-8}{4x-1}+\frac{2x}{x^2+1}+\frac{9}{x^2+1}dx\]

Answer 65

This one is a going to be a little bit painful to integrate lol.
We're going to get a bunch of different stuff :)

Answer 66

Yeah! This is a little strange to integrate!

Answer 67

What you got?

Answer 68

The first term isn't too bad.
It's going to give us another natural log.
Except we'll have an extra factor of 1/4 in front due to the 4 on our x.
If you're confused about that, then do a u-sub.

Answer 69

\[\Large -8\int\limits \frac{1}{4x-1}\;dx \quad=\quad -8\cdot\frac{1}{4}\ln|4x-1|\]
Confused by the 1/4? :o

Answer 70

No! I got it!

Answer 71

Well Thank you so much! I know it's late, and you are probably exhausted! I'll try to figure it out by my own!

Answer 72

You have helped a lot!

Answer 73

ok ok ok I'll just give ya some quick tips so you don't get too confused :)o
The second one is a simple u-sub (denominator is u);.
The third piece is one that you really really should have memorized.\[\Large \int\limits\frac{1}{1+x^2}dx \quad=\quad?\]
Remember that one? :)

Answer 74

Yes, I do! ;)