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Mathematics 13 Online
OpenStudy (konradzuse):

x^2/(x^2-1)

OpenStudy (konradzuse):

This is part of an integral by parts q, which apparently needs long division...

OpenStudy (anonymous):

Rearrange the numerator It will become simple then..

OpenStudy (konradzuse):

I got 1+1/(x^2-1) as the answer, but apparently it's 1- 1/2(x+1) +2/2(x-1)

OpenStudy (lgbasallote):

1 + 1/(x^2 - 1) is your quotient from long div?

OpenStudy (anonymous):

\[x^2 \implies x^2 - 1 + 1\]

OpenStudy (anonymous):

So it will become: \[\int\limits 1.dx + \int\limits \frac{1}{x^2 - 1}dx\]

OpenStudy (lgbasallote):

@waterineyes i think the book used PFs..

OpenStudy (lgbasallote):

dont you think so?

OpenStudy (konradzuse):

hmmm.....

OpenStudy (anonymous):

PF ??

OpenStudy (lgbasallote):

partial fractions

OpenStudy (anonymous):

There is a special formula for the last part..

OpenStudy (konradzuse):

well iggy I did x^2- 1 | x^2 and x^2/x^2 = 1 1 * x^2 = x^2 1 * -1 = -1..

OpenStudy (konradzuse):

since there was nothing else I did what your sheet said and did 1/x^2-1

OpenStudy (lgbasallote):

\[\frac{1}{x^2 - 1} = \frac{A}{x+1} + \frac{B}{x-1}\] are you familiar with that method?

OpenStudy (lgbasallote):

@KonradZuse you are right...but the book or whoever provided the answer took it a step more

OpenStudy (anonymous):

Yes I am.

OpenStudy (konradzuse):

Oic yeah...

OpenStudy (lgbasallote):

i was referring to konrad lol. i know you're familiar

OpenStudy (anonymous):

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