x^2/(x^2-1)
This is part of an integral by parts q, which apparently needs long division...
Rearrange the numerator It will become simple then..
I got 1+1/(x^2-1) as the answer, but apparently it's 1- 1/2(x+1) +2/2(x-1)
1 + 1/(x^2 - 1) is your quotient from long div?
\[x^2 \implies x^2 - 1 + 1\]
So it will become: \[\int\limits 1.dx + \int\limits \frac{1}{x^2 - 1}dx\]
@waterineyes i think the book used PFs..
dont you think so?
hmmm.....
PF ??
partial fractions
There is a special formula for the last part..
well iggy I did x^2- 1 | x^2 and x^2/x^2 = 1 1 * x^2 = x^2 1 * -1 = -1..
since there was nothing else I did what your sheet said and did 1/x^2-1
\[\frac{1}{x^2 - 1} = \frac{A}{x+1} + \frac{B}{x-1}\] are you familiar with that method?
@KonradZuse you are right...but the book or whoever provided the answer took it a step more
Yes I am.
Oic yeah...
i was referring to konrad lol. i know you're familiar
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