megannicole51
  • megannicole51
Split into partial fractions.....((64)/(64-x^2))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
megannicole51
  • megannicole51
I get that you have to factor out the denominator which gives you (8-x)(x+8)...but then what?
Loser66
  • 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
  • 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
  • megannicole51
actually i dont have it....im still confused
anonymous
  • 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
  • anonymous
and solve
megannicole51
  • megannicole51
so to find A i do.....A(8+-8)+B(8+x)=64
anonymous
  • anonymous
yes... you'll get two equations, one for the number/constant term and one for the x term
anonymous
  • anonymous
\[8A + 8B = 64 \text{, and } -Ax + Bx = 0x\]
anonymous
  • anonymous
Once you've found A & B then put them back into the partial fraction decomposition and integrate the terms separately.
megannicole51
  • megannicole51
so would A=4 as well?
megannicole51
  • megannicole51
I just plugged b=4 into 8A+8B=64
anonymous
  • anonymous
4? I think both A & B are 1...
megannicole51
  • megannicole51
the dude who was helping me before said b=4?
anonymous
  • anonymous
yeah, sorry. Both A & B are 4.
megannicole51
  • megannicole51
yay okay! so then to split up the equations they would be ((4)/(8-x)) and ((4)/(8+x))?
anonymous
  • anonymous
so you get \[ \int \frac{64}{64-x^{2}}\, dx =\int \left(\frac{4}{8-x} + \frac{4}{8-x}\right) dx \]
anonymous
  • anonymous
an you integrate those?
anonymous
  • anonymous
can...
megannicole51
  • megannicole51
no
megannicole51
  • megannicole51
and shouldnt one of them be 8+x in the numerator?
anonymous
  • 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
  • megannicole51
okay:) thank you my notes are so confusing
anonymous
  • 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
  • megannicole51
oooh u substitution! okay yeah that makes sense!
anonymous
  • anonymous
it's pretty helpful, that one!
megannicole51
  • megannicole51
wait so how is that the integration of 64/64-x^2?
anonymous
  • anonymous
that's only the first part... you need to do the second integral, too.
anonymous
  • anonymous
\[\int \frac{4}{8+x}\, dx\]
megannicole51
  • megannicole51
so would it be the same as the first one except its (8+x)?
anonymous
  • 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...
megannicole51
  • megannicole51
oooh okay so it would be 4ln(8+x)+c
anonymous
  • anonymous
here's a paper that has it all (for partial fraction decomposition and integration)...
anonymous
  • anonymous
yes!
megannicole51
  • megannicole51
these papers are awesome thank you so much! and so then do you just add each integration?
anonymous
  • 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
  • anonymous
you're welcome... thanks for being such a great learner!
megannicole51
  • megannicole51
take the derivative of what sum?
megannicole51
  • megannicole51
well thanks for being such a great teacher!
anonymous
  • anonymous
\[ 4\ln|8+x| - 4\ln|8-x|\]
anonymous
  • anonymous
I forgot the absolute values for the indefinite integral yielding ln.
megannicole51
  • megannicole51
why is it the absolute value now?
anonymous
  • 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
  • anonymous
so much to remember!!!
megannicole51
  • megannicole51
so don't they just cancel each other out and ur just left with c?
megannicole51
  • megannicole51
and i know right!!
anonymous
  • anonymous
no, the arguments are not the same
anonymous
  • 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
  • megannicole51
wait so isnt that what we started with? and that makes sense but idk why the absolute values are necessary
anonymous
  • 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
  • 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
  • megannicole51
ooh okay that makes sense. so the answer to the integration of 64/64-x^2 is what u just wrote?
anonymous
  • anonymous
the derivative of the constant is, of course, 0.
megannicole51
  • megannicole51
and then +c at the end of it?
anonymous
  • anonymous
yeah... \[ \int \frac{64}{64-x^{2}}\,dx = 4\ln|8+x|-4\ln|8-x|+C\]
anonymous
  • anonymous
you good?
megannicole51
  • 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
  • anonymous
shift \ is an upright bar (|). maybe your online program doesn't require it?
megannicole51
  • megannicole51
i tried just putting parenthesis and it didnt work
anonymous
  • anonymous
did you try | ?
megannicole51
  • megannicole51
yeah it didnt work:/
anonymous
  • 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
  • megannicole51
i emailed my professor i guess we will see....thanks for your help!
anonymous
  • anonymous
you're welcome!

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