Question

I have a semi time consuming one i need help with... just a heads up! i haven't seen this in a while so i'm super rusty :(
integral

Answer 1

Oil can?

Answer 2

\[\int\limits_{:(}^{:)} \frac{ 1 }{ x ^{2} -1}dx\]

Answer 3

oops! the bottom is squared

Answer 4

\[\int\limits_{:)}^{o.0}\frac{ 1 }{ (x ^{2}-1)^{2}}dx\]

Answer 5

Are those supposed to be "Infinity" symbols? \(\infty\)?

Have you considered Partial Fractions? It will be fun!

Answer 6

they are just faces. there are no bounds!

Answer 7

Ah, so just the indefinite integral. No more faces.

Answer 8

yeah. i know this is the way to go, i just cant remember how to put them together! i know i look for the degree and A/something +B/something^2 + c/something + d/something^2

Answer 9

ooh! i tried to erase the question mark and it redid the symbol into words, maybe i just did it wrong. sorry!

Answer 10

it's the part after that i get a bit lost

Answer 11

That's it. Factor Completely. \((x^{2} - 1)^{2} = (x+1)^{2}(x-1)^{2}\) and there are your "something" and "something squared" things. \((x+1), (x+1)^{2}, (x-1), (x-1)^{2}\)

Answer 12

yeah, i think it's the putting the 1 = ( )( )A + ( ) ( )B + ( )( )C + ( )D where i am lost. ok here's where i am...
\[\frac{ A }{ x+2 }+\frac{ B }{ (x+2)^{2} } + \frac{ C }{ x-2 } + \frac{ D }{ (x-2)^{2} }\]

Answer 13

Whence came all the 2's? Unless you provided the wrong problem statement, those should all be 1's.

Answer 14

oops you're right. i wrote it down wrong on this paper. been a long day! but yeah, those are meant to be 1's

Answer 15

Okay, now do the algebra.

\(A(x+1)(x-1)^{2} + B(x-1)^{2} + ...\)

Answer 16

i think it's this one algebra step i dont understand

Answer 17

We're just finding a Common Denominator. You've done that lots of times.

Answer 18

oooh.ha! you're right. I think my teacher just explained it a bit differently with a smaller one and threw me off

Answer 19

Do the algebra. I'll give you one free mercy answer.

A = 1/4
B = 1/4
C = -1/4
D = 1/4

Make that happen!