Question

Split into partial fractions.....((64)/(64-x^2))

Answer 1

I get that you have to factor out the denominator which gives you (8-x)(x+8)...but then what?

Answer 2

\[\frac{64}{(8+x)(8-x)}=\frac{A}{(8+x)}+\frac{B}{(8-x)}\]
therefore, A (8-x) + B (8+x) = 64
when x =8 \(\rightarrow \)A (8-x) =0 and B (8+8) = 16 B = 64 \(\rightarrow\) B = 4
when x = -8 \(\rightarrow\) immitate my step above to find out A,
then plug A and B back to the partial fraction above, you have the answer

Answer 3

got it! thank you:) so then how do u finish the integral after that? I have to split it up and then solve it

Answer 4

actually i dont have it....im still confused

Answer 5

you have to multiply to get the common denominator under each term on the right hand side (rhs). after that, you set each part equivalent.

Answer 6

and solve

Answer 7

so to find A i do.....A(8+-8)+B(8+x)=64

Answer 8

yes... you'll get two equations, one for the number/constant term and one for the x term

Answer 9

\[8A + 8B = 64 \text{, and } -Ax + Bx = 0x\]

Answer 10

Once you've found A & B then put them back into the partial fraction decomposition and integrate the terms separately.

Answer 11

so would A=4 as well?

Answer 12

I just plugged b=4 into 8A+8B=64

Answer 13

4? I think both A & B are 1...

Answer 14

the dude who was helping me before said b=4?

Answer 15

yeah, sorry.
Both A & B are 4.

Answer 16

yay okay! so then to split up the equations they would be ((4)/(8-x)) and ((4)/(8+x))?

Answer 17

so you get

\[ \int \frac{64}{64-x^{2}}\, dx =\int \left(\frac{4}{8-x} + \frac{4}{8-x}\right) dx \]

Answer 18

an you integrate those?

Answer 19

can...

Answer 20

no

Answer 21

and shouldnt one of them be 8+x in the numerator?

Answer 22

no... the decomposition written above is what you want...

\[\int \frac{4}{8-x}\, dx + \int \frac{4}{8+x}\, dx \]

let's do this first one together...

Answer 23

okay:) thank you

my notes are so confusing

Answer 24

\[\int \frac{4}{8-x}\, dx \text{ Let } u=8-x \text{, } du = -dx \text{. Then} \]
\[ \int \frac{4}{8-x}\, dx = -4\int \frac{-dx}{8-x} = -4\int \frac{du}{u} = -4\ln u + C = -4\ln(8-x) +C\]

Answer 25

oooh u substitution! okay yeah that makes sense!

Answer 26

it's pretty helpful, that one!

Answer 27

wait so how is that the integration of 64/64-x^2?

Answer 28

that's only the first part... you need to do the second integral, too.

Answer 29

\[\int \frac{4}{8+x}\, dx\]

Answer 30

so would it be the same as the first one except its (8+x)?

Answer 31

No... no negative.
Just practice doing it all until you get comfortable with it all. Otherwise, it's easy to make mistakes and you won't have a solid methodology to fall back on when things get tough or don't work out. Know what I mean?

here's a paper on what to do when you have an irreducible quadratic witth 3 terms...

Answer 32

oooh okay so it would be 4ln(8+x)+c

Answer 33

here's a paper that has it all (for partial fraction decomposition and integration)...

Answer 34

yes!

Answer 35

these papers are awesome thank you so much! and so then do you just add each integration?

Answer 36

take the derivative of that sum and you'll see you get what we integrated, which is equivalent to the original problem statement.

Answer 37

you're welcome... thanks for being such a great learner!

Answer 38

take the derivative of what sum?

Answer 39

well thanks for being such a great teacher!

Answer 40

\[ 4\ln|8+x| - 4\ln|8-x|\]

Answer 41

I forgot the absolute values for the indefinite integral yielding ln.

Answer 42

why is it the absolute value now?

Answer 43

because of the domain of the ln function. if we had limits of integration, we'd need to make sure whatever we put in the ln function was > 0

Answer 44

so much to remember!!!

Answer 45

so don't they just cancel each other out and ur just left with c?

Answer 46

and i know right!!

Answer 47

no, the arguments are not the same

Answer 48

\[ \frac{d}{dx} \left( 4\ln|8+x| - 4\ln|8-x| \right) = \frac{4}{8+x} - \frac{-4}{8-x}= \frac{4}{8+x} + \frac{4}{8-x}\]

Answer 49

wait so isnt that what we started with? and that makes sense but idk why the absolute values are necessary

Answer 50

of course it's what we started with! integration is anti-differentiation so it undoes taking the derivative. likewise, once you've integrated, you can take the derivative to check that you get the same thing you just integrated. that let's you know that you've done the integration correctly.
As for the absolute values... if x = 9 then ln(8-x) = ln(8-9) = ln(-1) which isn't allowed because the domain of ln y is y>0. here. y = 8-x so 8-x>0 => x<8.
Does that make sense? The absolute value is just a precautious technicallity to prevent putting invalid values in the function. the absolute value ensures the argument of the natural log is positive.

Answer 51

oh, and sorry but it technically should have been this...

\[ \frac{d}{dx}\left(4\ln(8+x)-4\ln(x-8)+C\right)=\frac{4}{8+x} + \frac{4}{8-x} \]

Answer 52

ooh okay that makes sense. so the answer to the integration of 64/64-x^2 is what u just wrote?

Answer 53

the derivative of the constant is, of course, 0.

Answer 54

and then +c at the end of it?

Answer 55

yeah...
\[ \int \frac{64}{64-x^{2}}\,dx = 4\ln|8+x|-4\ln|8-x|+C\]

Answer 56

you good?

Answer 57

is there another way to write the absolute value? i tried entering it into my online homework stuff and idk how to put the absolute value

Answer 58

shift \ is an upright bar (|). maybe your online program doesn't require it?

Answer 59

i tried just putting parenthesis and it didnt work

Answer 60

did you try | ?

Answer 61

yeah it didnt work:/

Answer 62

i don't what to say... maybe ask your teacher or someone in your class how to type absolute values in your program. does the program have a help section? it may tell you how to do that.

Answer 63

i emailed my professor i guess we will see....thanks for your help!

Answer 64

you're welcome!