Express following in partial fractions...evaluate the given integral

- anonymous

Express following in partial fractions...evaluate the given integral

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- anonymous

|dw:1438068450403:dw|

- anonymous

@UsukiDoll

- UsukiDoll

oh crap... let me expand (x-1)^2(x+1) first with wolfram and see if it comes back as factorable all the way.

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## More answers

- UsukiDoll

ok it's all factorable so we have single letters to deal with.. one problem whenever we have a repeater... we have to take the original and build up the exponents.

- UsukiDoll

for example |dw:1438068733033:dw| then take lcd of all of these guys

- UsukiDoll

luckily the problem we are given isn't harsh like that (though I did s^5 at one point)

- UsukiDoll

show me your attempt :)

- anonymous

ok hold on

- anonymous

\[\frac{ 2x }{ (x-2)^2(x+1) }=\frac{ A }{ x+1 }+\frac{ B }{(x-2)^2 }\]
simplify
\[2x=\frac{ A(x+2 )((x-2)^2) }{ x+2 } + \frac{ B((x-2)^2(x+2)) }{ (x-2)^2 }\]

- anonymous

and then

- UsukiDoll

STOP!

- UsukiDoll

the A part is fine.. the one with the exponent.. no...

- anonymous

\[2x=A(x-2)^2+B(x+1)\]
x=2 gives
\[4=A(0)^2+B(3)\]
B=4/3

- anonymous

ohh

- UsukiDoll

whenever you have the exponent .. you have to write the single version first and then the exponent (depending on the number) look at this example
|dw:1438069380011:dw|

- UsukiDoll

I wrote the single version s and then work the exponent up
s s^2 s^3 s^4
then you got to take lcd of all those fractions

- UsukiDoll

|dw:1438069450631:dw|

- UsukiDoll

I think order matters too... |dw:1438069483088:dw|

- UsukiDoll

the lcd for this is \[(x+1)(x-1)^2 \]
alright take the lcd ;)

- UsukiDoll

ahem... take the lcd for |dw:1438069618368:dw|

- anonymous

alright alright, give me a sec:)

- anonymous

|dw:1438069792243:dw|

- UsukiDoll

stop setting 2x =
that's not going to work

- UsukiDoll

the lcd part is good though

- UsukiDoll

|dw:1438069902990:dw|

- UsukiDoll

k expand all of that

- UsukiDoll

*eats popcorn while I wait*

- anonymous

I dont get this at all....whats wrong with me!!

- UsukiDoll

oyyyyyyyy........

- UsukiDoll

|dw:1438070102575:dw|

- UsukiDoll

your turn :)

- anonymous

whats does foil mean':)?

- UsukiDoll

foil - first outer inner last... okkk expand :)

- anonymous

Maybe I shouldve told u havent slept in 2 days...

- UsukiDoll

then you should take a break from math and sleep

- anonymous

its 10am...my exam is in 1 month from today....I have alot to do...due to circumstances...I cant sleep til 12hours from now,

- UsukiDoll

try finishing the problem?

- anonymous

so expanding c gives
x^2-2x+1

- anonymous

for A gives x^2-1

- UsukiDoll

|dw:1438070573305:dw|
|dw:1438070618427:dw|

- UsukiDoll

yeah...

- UsukiDoll

|dw:1438070677726:dw|

- UsukiDoll

now distribute the letters

- anonymous

Im really sry, can you finish it?
I cant think properly, after this im going to eat and take a break and then start this again fresh.

- UsukiDoll

the problem is that I can't give direct answers... so can't finish it without getting busted.

- anonymous

@aric200

- UsukiDoll

@Xlegalize can you come back over here so we can finish this please?

- anonymous

im here

- UsukiDoll

ok so the last step we left off is |dw:1438073386251:dw|
so what we need to do next is to distribute the letters

- UsukiDoll

I'll give an example
\[A(x^2-1) = Ax^2-A \]

- UsukiDoll

then similarly for B and C
\[B(x+1) \rightarrow Bx+B \]
\[C(x^2-2x+1) \rightarrow Cx^2-C2x+C \]

- UsukiDoll

yeahhh that 2xC combination looks gross... and awkward XD anyway now we can put that long string in the numerator .

- UsukiDoll

\[\large \frac{Ax^2-A+Bx+B+Cx^2-2xC+C}{(x+1)(x-1)^2}\]

- UsukiDoll

|dw:1438073620766:dw| on the original equation, we have a 2x so that would mean that all equations with x would be equal to 2. whereas the plain letters and the letters with x^2 will be set equal to 0

- UsukiDoll

\[-A+B+C =0\]
\[B-2C=2 \]
\[A+C =0 \]

- UsukiDoll

so now we have these systems of equations and we need to solve for A,B,C
once we get A B C we can plug it back into the equation from earlier

- anonymous

B=2+2c
A=2+2c+c
A+c=0
2+2c+c+c=0

- UsukiDoll

let's try
adding the first and third equation up
\[-A+B+C =0 \]
\[A+C =0\]
the A cancels out and we are left with
B+2C= 0

- anonymous

ok that is easier.

- UsukiDoll

so now adding
B-2C=2
B+2C=0
we have
2B=2
B = 1

- UsukiDoll

so now that B = 1 the only place I can put that is the second equation
if I put this in the first equation I'm stuck.. because I don't know what C or A is...yet

- UsukiDoll

B-2C=2
1-2C=2
|dw:1438074079535:dw|

- UsukiDoll

so B = 1, C = -1/2
now we can get our A
picking out an easier equation with A is the best method to getting what A is

- anonymous

a=1/2

- UsukiDoll

|dw:1438074167001:dw|

- UsukiDoll

that's right..
now that A = 1/2, B = 1, and C = -1/2
we need to check if they satisfy the three equations. If they do then we can go back, plug and chug into the earlier equation, and use integration

- UsukiDoll

\[A+C =0 \] since A = 1/2 and C = -1/2
\[\frac{1}{2}-\frac{1}{2}=0 \]
\[0=0 \]
\[-A+B+C =0 \]
\[-\frac{1}{2}+1-\frac{1}{2} =0\]
\[-\frac{1}{2}-\frac{1}{2}+1 =0\]
\[-\frac{1}{2}-\frac{1}{2}+1 =0\]
\[-\frac{2}{2}+1 =0\]
\[-1+1 =0\]
\[0 =0\]
\[B-2C=2 \]
B = 1
\[1-2\frac{-1}{2}=2 \]
\[1+\frac{2}{2}=2 \]
\[1+1=2 \]
\[2=2 \]
ok after that long verification latex string we can plug our values into the earlier equation and integrate

- UsukiDoll

|dw:1438074628205:dw|

- UsukiDoll

so after all that lcd solving for a b c we can finally integrate... YAY! 2 of them are logs.. one is u substitution

- UsukiDoll

|dw:1438074757028:dw| so the red portion is the \[\frac{1}{(x-1)^2}\] part. we have to do a bit more work on it

- anonymous

Go on:)

- anonymous

yeah I also got those parts.

- anonymous

the last one takes more work.

- UsukiDoll

yeah x.x now.. using u sub.
u = x-1
du = 1 dx

- UsukiDoll

|dw:1438074892358:dw| I wanna recheck that just to be sure

- anonymous

yep thats correct.

- UsukiDoll

thanks :)
so final answer after all that jazz is
|dw:1438075034959:dw|

- anonymous

I will take a break and go through this all again. So Im 100 on this
THANKS!!!!

- UsukiDoll

you're welcome :)

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