Question

Partial Fraction Problem..

Answer 1

\[\int\limits_{-1}^{1} \frac{ 1 }{ (x+5)(x^2+9) }\]

Answer 2

Decomposition? c:
Oh these are kinda fun.

Answer 3

yes! i enjoy this too.. but i got stuck. so I started with A/(x+5) + B/(x^2+9)

Answer 4

Woops!

Answer 5

The numerator should always be `one degree less than the denominator`

Answer 6

See how your second factor is quadratic?

Answer 7

So the numerator needs to be `linear`

Answer 8

yes. so do i need to factor it?

Answer 9

\[\large\rm \frac{1}{(x+5)(x^2+9)}\quad=\quad \frac{A}{x+5}+\frac{Bx+C}{x^2+9}\]

Answer 10

Unfortunately, since it's the `sum of squares`, it won't factor in any useful way.

Answer 11

If it was `difference of squares`, yes factoring would be nice :)

Answer 12

But anyway, that's what your initial setup should look like.
Numerator a lil confusing? :o

Answer 13

oh ok.. i haven't done it this way before! the Bx + C is confusing not sure what i would do next? is it A(x^2+9) + Bx+C(x+5) ?

Answer 14

Multiplying through by the denominator on the left side gives us,\[\large\rm 1=A(x^2+9)+(Bx+C)(x+5)\]Yes, but don't forget the brackets around your Bx+C.
The (x+5) is multiplying the entire thing.

Answer 15

I got
\[Ax^2+9A+Bx^2+C+5Bx+C\]

Answer 16

Oh you like to expand it all out?
Do it that way?
Barf :P ok fine we can do that.

Answer 17

oh i don't know... lol if there is an easier way ill do it !

Answer 18

It's `sometimes` easier.
Just depends on the problem.

But it's probably good to have both methods in your goodie bag.

Answer 19

The other way to approach this is...
We just pick clever values for x to figure out our A, B and C.

For example, x=-5 will allow us to solve for A:\[\large\rm 1=A((-5)^2+9)+(-5B+C)(-5+5)\]\[\large\rm 1=A(25+9)+(-5B+C)(0)\]\[\large\rm 1=34A\]\[\large\rm A=\frac{1}{34}\]

Answer 20

oh ok! can I always do it this way for partial fractions?

Answer 21

The values to pick are the ones that will (ideally) cancel out all but one constant. In the cases where this is not possible, you''ll have to make linear equations and "separate" the constants there.

Answer 22

oh ok i see! thank you

Answer 23

Always? Umm..
I suppose, no.
If you have a bunch of squares in the denominator that can't factor down further,
then you might run into a situation where you can't cancel out some of the factors.
But then it just turns into the other method, system of equations, at that point.

Answer 24

Btw, Great explanations @zepdrix!

Answer 25

But it usually works :)
So next let's try another cool value.
x=0 is going to be very helpful.

Answer 26

Oh before we do that, let's make sure we've plugged in our new-found A value.

Answer 27

ok. so i plug in 1/34 into the equation

Answer 28

into A

Answer 29

\[\large\rm 1=\frac{1}{34}(x^2+9)+(Bx+C)(x+5)\]And then x=0,\[\large\rm 1=\frac{1}{34}(0^2+9)+(B\cdot0+C)(0+5)\]So which variable is x=0 going to allow us to find?

Answer 30

C

Answer 31

Yah, the B is multiplying 0, so C.
Find your C!
Lemme know what you get! :)

Answer 32

so far i have C = 1 - 45/34. is that correct to continue /?

Answer 33

Hmm that doesn't look quite right :p thinking..

Answer 34

From here,\[\large\rm 1=\frac{1}{34}(0^2+9)+(B\cdot0+C)(0+5)\]Getting rid of all the 0's shows us that we have,\[\large\rm 1=\frac{9}{34}+5C\]

Answer 35

Subtract 9/34 to the other side, ya?

Answer 36

Get a common denominator and subtract the 9/34 from the 1,
do that before you do anything with the 5.

Answer 37

ok!

Answer 38

Comeon lil giraffe >.< you can do it!
Gimme that C!

Answer 39

hmm i got c= -30/170 ....

Answer 40

Hmm :d

Answer 41

wait sorry

Answer 42

25/170 ?

Answer 43

c= 5/34 ?

Answer 44

k good!

Answer 45

yay!

Answer 46

\[\large\rm 1=\frac{1}{34}(x^2+9)+\left(Bx+\frac{5}{34}\right)(x+5)\]Umm this might be crazy, but let's multiply both sides by 34,
it just might make it easier to work with.\[\large\rm 34=(x^2+9)+\left(34Bx+5\right)(x+5)\]

Answer 47

And now at this point, we have a single unknown, B.
So we can plug in pretty much any value for x to solve for B.
But let's be careful not to choose a number that we've already used.

How about x=1, that seems simple enough.

Answer 48

\[\large\rm 34=(1^2+9)+\left(34B\cdot1+5\right)(1+5)\]We ok up to this point? :)
Or still trying to wrap your head around that 34 multiplication? lol

Answer 49

i think i got it you multiplied a number that would get rid of the denominator to make it nicer. and you chose 1 to plug in

Answer 50

so i have 34 = 10 + (B+5) + (6) ?

Answer 51

oops i forgot the 34

Answer 52

Ya the 34, but also the 6 is multiplying, not adding, ya?\[\large\rm 34=10+\left(34B+5\right)(6)\]

Answer 53

oh yes!

Answer 54

so i have 34 = 10+204B+30

Answer 55

k cool.
solve for B.

Answer 56

b = 34/244?

Answer 57

mmmmm no

Answer 58

\[\large\rm 34=10+\left(34B+5\right)(6)\]Subtract 10,\[\large\rm 24=\left(34B+5\right)(6)\]At this point, if you like to distribute the 6, that's fine.
I don't like the big numbers that it produces though,
so I would opt to divide both sides by 6 at this point.\[\large\rm 4=34B+5\]Doing it your way will work out fine though.
You'll just have to simplify the fraction at some point.

Answer 59

That wasn't the problem though, you made some algebra mistake in there.
I'm not sure where though :d

Answer 60

ohh ok i see so i get B= -1/34

Answer 61

Yay!

Answer 62

now i put it in the integral and solve

Answer 63

\[\rm \int\limits\limits_{-1}^{1} \frac{1}{(x+5)(x^2+9)}\quad=\quad \int\limits_{-1}^{1}\frac{1/34}{x+5}+\frac{(-1/34)B+(5/34)}{x^2+9}~dx\]Yes :)
Let's apply another clever trick.
Let's factor a 1/34 out of each term, and it bring it to the front of the integral.

Answer 64

Woops that should be (-1/34)x not (-1/34)B *

Answer 65

\[\large\rm =\quad\frac{1}{34}\int\limits_{-1}^{1}\frac{1}{x+5}+\frac{-x+5}{x^2+9}~dx\]

Answer 66

so i can separate into 2 integrals

Answer 67

the first will be equal to ln|(x+5)|?

Answer 68

You'll need to split that second fraction up,
so you'll actually have 3 integrals.\[\large\rm =\quad\frac{1}{34}\int\limits\limits_{-1}^{1}\frac{1}{x+5}+\frac{-x}{x^2+9}+\frac{5}{x^2+9}~dx\]

Answer 69

First one looks good.

Answer 70

Try to give those other two a try.
Imma go make a snack real quick :)

Answer 71

lol ok thank you!

Answer 72

for the second integral i did u = x^2+9 and 1/2du=xdx

Answer 73

i got -1/u as my integral

Answer 74

Pulling the 1/2 out front?

Answer 75

yes

Answer 76

i have it behind the integral..

Answer 77

left side is the front :3
if that's what you mean XD

Answer 78

ohh ok yes!

Answer 79

thats what i mean

Answer 80

i dunno why they call that side the front :P
probably because we read from left to right, so front comes first.

Answer 81

lol ya! so that integral is easy to solve i can make it -u^-1. but the other integral I'm having trouble with

Answer 82

ya, this is why we had to split it up.
That last one ends up being an inverse trig.
requires a trig sub.

Answer 83

Just a note you might want to make.
When you're given a denominator with `only linear factors`, example:\[\large\rm \int\limits\frac{1}{(x+3)(x-2)(x+2)}dx\]in that case, the method we used,
plugging in specific values for x,
will always be better than expanding.
At least in my opinion.

When we have a square that can't be broken down:\[\large\rm \int\limits \frac{1}{(x+5)(x^2+9)}dx\]Then either method is probably about the same amount of work.
Use which ever one you like.

Answer 84

\[\large\rm \int\limits\frac{5}{x^2+9}dx\]So we need to think about which Pythagorean Identity has (something)^2+1.
Which one is it?
The sine identity?
The tangent identity?

Hmm which one?

Answer 85

oh i know.. x=atan theta?

Answer 86

yah

Answer 87

thank you! i think i got it from now:) it will take me a while to do it but i am good with trig sub! So I'm sure i can handle it! thanks so much!!!!!!!

Answer 88

np ୧ʕ•̀ᴥ•́ʔ୨