calculus 2 - partial fractions
\[\frac{a}{4-x}+\frac{b}{4+x}\] solve for \(a,b\)
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?
wanna do it with the 40 or without?
no i don't. Whichever is the best and quickest way in your opinion
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 \(A=5\)
shouldn't that not work since you are placing 4 in x for both sides?
A/ (4-4) is undefined?
that is why you cover it up with your finger when you compute
i was only plugging 4 in on the left
is that legal in math?
\[A=\frac{40}{\cancel{(4-x})(4+4)}=\frac{40}{8}=5\]
it is just a quick technique you accomplish the same thing by writing \[40=A(4+x)+B(4-x)\] and then saying "let \(x=4\)"
Ok, I get it. You have to times (4-x) on each side first to cancel them out.
you would get \[40=8A\] and so \(A=5\) same idea exactly
just a quick gimmick is all, but it works nicely btw i thing \(B=5\) as well
*think
ok cool. what to do from there?
\[\int \frac{5}{4-x}dx+\int \frac{5}{4+x}dx\] should be immediate right?
i think so
i am wondering if i made a mistake somewhere
no, i am sticking with my answer
What would the final answer be?
well my problem is that wolfram gave me a different answer, but i like my answer of \[5\ln(4-x)+5\ln(4+x)\]
i tried the 3rd and fifth bubble. both are wrong
i get 3, but wolfram gets 4
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
should i try the 4th bubble? This is my last guess
wolf get \[5\ln(\frac{4+x}{4-x})\]
That got me the right answer
i still like my answer take the derivative and it works
thank you
yw
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