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Mathematics 10 Online
OpenStudy (anonymous):

calculus 2 - partial fractions

OpenStudy (anonymous):

OpenStudy (anonymous):

\[\frac{a}{4-x}+\frac{b}{4+x}\] solve for \(a,b\)

OpenStudy (anonymous):

you can ignore the 40 if you like and pull it out of the integral or leave it in there is a real quick snappy way to do this do you know it?

OpenStudy (anonymous):

wanna do it with the 40 or without?

OpenStudy (anonymous):

no i don't. Whichever is the best and quickest way in your opinion

OpenStudy (anonymous):

lets leave it in start with \[\frac{40}{(4-x)(4+x)}=\frac{A}{4-x}+\frac{B}{4+x}\] then \(4-x\) is 0 if \(x=4\) so put your finger over the factor of \(4-x\) in the original expression and replace \(x\) by \(4\) \[\frac{40}{\cancel{(4-x})(4+4)}=\frac{40}{8}=5\] and so \(A=5\)

OpenStudy (anonymous):

shouldn't that not work since you are placing 4 in x for both sides?

OpenStudy (anonymous):

A/ (4-4) is undefined?

OpenStudy (anonymous):

that is why you cover it up with your finger when you compute

OpenStudy (anonymous):

i was only plugging 4 in on the left

OpenStudy (anonymous):

is that legal in math?

OpenStudy (anonymous):

\[A=\frac{40}{\cancel{(4-x})(4+4)}=\frac{40}{8}=5\]

OpenStudy (anonymous):

it is just a quick technique you accomplish the same thing by writing \[40=A(4+x)+B(4-x)\] and then saying "let \(x=4\)"

OpenStudy (anonymous):

Ok, I get it. You have to times (4-x) on each side first to cancel them out.

OpenStudy (anonymous):

you would get \[40=8A\] and so \(A=5\) same idea exactly

OpenStudy (anonymous):

just a quick gimmick is all, but it works nicely btw i thing \(B=5\) as well

OpenStudy (anonymous):

*think

OpenStudy (anonymous):

ok cool. what to do from there?

OpenStudy (anonymous):

\[\int \frac{5}{4-x}dx+\int \frac{5}{4+x}dx\] should be immediate right?

OpenStudy (anonymous):

i think so

OpenStudy (anonymous):

i am wondering if i made a mistake somewhere

OpenStudy (anonymous):

no, i am sticking with my answer

OpenStudy (anonymous):

What would the final answer be?

OpenStudy (anonymous):

well my problem is that wolfram gave me a different answer, but i like my answer of \[5\ln(4-x)+5\ln(4+x)\]

OpenStudy (anonymous):

i tried the 3rd and fifth bubble. both are wrong

OpenStudy (anonymous):

i get 3, but wolfram gets 4

OpenStudy (anonymous):

on the other hand, if you take the derivative of number 3 you get \(\frac{40}{16-x^2}\) so i am not sure what is wrong

OpenStudy (anonymous):

should i try the 4th bubble? This is my last guess

OpenStudy (anonymous):

wolf get \[5\ln(\frac{4+x}{4-x})\]

OpenStudy (anonymous):

That got me the right answer

OpenStudy (anonymous):

i still like my answer take the derivative and it works

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

yw

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