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

- megannicole51

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

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- schrodinger

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- megannicole51

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

- Loser66

\[\frac{64}{(8+x)(8-x)}=\frac{A}{(8+x)}+\frac{B}{(8-x)}\]
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

- megannicole51

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

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- megannicole51

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

- anonymous

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.

- anonymous

and solve

- megannicole51

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

- anonymous

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

- anonymous

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

- anonymous

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

- megannicole51

so would A=4 as well?

- megannicole51

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

- anonymous

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

- megannicole51

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

- anonymous

yeah, sorry.
Both A & B are 4.

- megannicole51

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

- anonymous

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

- anonymous

an you integrate those?

- anonymous

can...

- megannicole51

no

- megannicole51

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

- anonymous

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...

- megannicole51

okay:) thank you
my notes are so confusing

- anonymous

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

- megannicole51

oooh u substitution! okay yeah that makes sense!

- anonymous

it's pretty helpful, that one!

- megannicole51

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

- anonymous

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

- anonymous

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

- megannicole51

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

- anonymous

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...

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- megannicole51

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

- anonymous

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

##### 1 Attachment

- anonymous

yes!

- megannicole51

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

- anonymous

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

- anonymous

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

- megannicole51

take the derivative of what sum?

- megannicole51

well thanks for being such a great teacher!

- anonymous

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

- anonymous

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

- megannicole51

why is it the absolute value now?

- anonymous

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

- anonymous

so much to remember!!!

- megannicole51

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

- megannicole51

and i know right!!

- anonymous

no, the arguments are not the same

- anonymous

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

- megannicole51

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

- anonymous

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.

- anonymous

oh, and sorry but it technically should have been this...
\[ \frac{d}{dx}\left(4\ln(8+x)-4\ln(x-8)+C\right)=\frac{4}{8+x} + \frac{4}{8-x} \]

- megannicole51

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

- anonymous

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

- megannicole51

and then +c at the end of it?

- anonymous

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

- anonymous

you good?

- megannicole51

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

- anonymous

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

- megannicole51

i tried just putting parenthesis and it didnt work

- anonymous

did you try | ?

- megannicole51

yeah it didnt work:/

- anonymous

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.

- megannicole51

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

- anonymous

you're welcome!

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