anonymous
  • anonymous
Express following in partial fractions...evaluate the given integral
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1438068450403:dw|
anonymous
  • anonymous
@UsukiDoll
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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UsukiDoll
  • 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
  • UsukiDoll
for example |dw:1438068733033:dw| then take lcd of all of these guys
UsukiDoll
  • UsukiDoll
luckily the problem we are given isn't harsh like that (though I did s^5 at one point)
UsukiDoll
  • UsukiDoll
show me your attempt :)
anonymous
  • anonymous
ok hold on
anonymous
  • 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
  • anonymous
and then
UsukiDoll
  • UsukiDoll
STOP!
UsukiDoll
  • UsukiDoll
the A part is fine.. the one with the exponent.. no...
anonymous
  • anonymous
\[2x=A(x-2)^2+B(x+1)\] x=2 gives \[4=A(0)^2+B(3)\] B=4/3
anonymous
  • anonymous
ohh
UsukiDoll
  • 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
  • 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
  • UsukiDoll
|dw:1438069450631:dw|
UsukiDoll
  • UsukiDoll
I think order matters too... |dw:1438069483088:dw|
UsukiDoll
  • UsukiDoll
the lcd for this is \[(x+1)(x-1)^2 \] alright take the lcd ;)
UsukiDoll
  • UsukiDoll
ahem... take the lcd for |dw:1438069618368:dw|
anonymous
  • anonymous
alright alright, give me a sec:)
anonymous
  • anonymous
|dw:1438069792243:dw|
UsukiDoll
  • UsukiDoll
stop setting 2x = that's not going to work
UsukiDoll
  • UsukiDoll
the lcd part is good though
UsukiDoll
  • UsukiDoll
|dw:1438069902990:dw|
UsukiDoll
  • UsukiDoll
k expand all of that
UsukiDoll
  • UsukiDoll
*eats popcorn while I wait*
anonymous
  • anonymous
I dont get this at all....whats wrong with me!!
UsukiDoll
  • UsukiDoll
oyyyyyyyy........
UsukiDoll
  • UsukiDoll
|dw:1438070102575:dw|
UsukiDoll
  • UsukiDoll
your turn :)
anonymous
  • anonymous
whats does foil mean':)?
UsukiDoll
  • UsukiDoll
foil - first outer inner last... okkk expand :)
anonymous
  • anonymous
Maybe I shouldve told u havent slept in 2 days...
UsukiDoll
  • UsukiDoll
then you should take a break from math and sleep
anonymous
  • 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
  • UsukiDoll
try finishing the problem?
anonymous
  • anonymous
so expanding c gives x^2-2x+1
anonymous
  • anonymous
for A gives x^2-1
UsukiDoll
  • UsukiDoll
|dw:1438070573305:dw| |dw:1438070618427:dw|
UsukiDoll
  • UsukiDoll
yeah...
UsukiDoll
  • UsukiDoll
|dw:1438070677726:dw|
UsukiDoll
  • UsukiDoll
now distribute the letters
anonymous
  • 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
  • UsukiDoll
the problem is that I can't give direct answers... so can't finish it without getting busted.
anonymous
  • anonymous
@aric200
UsukiDoll
  • UsukiDoll
@Xlegalize can you come back over here so we can finish this please?
anonymous
  • anonymous
im here
UsukiDoll
  • 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
  • UsukiDoll
I'll give an example \[A(x^2-1) = Ax^2-A \]
UsukiDoll
  • UsukiDoll
then similarly for B and C \[B(x+1) \rightarrow Bx+B \] \[C(x^2-2x+1) \rightarrow Cx^2-C2x+C \]
UsukiDoll
  • UsukiDoll
yeahhh that 2xC combination looks gross... and awkward XD anyway now we can put that long string in the numerator .
UsukiDoll
  • UsukiDoll
\[\large \frac{Ax^2-A+Bx+B+Cx^2-2xC+C}{(x+1)(x-1)^2}\]
UsukiDoll
  • 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
  • UsukiDoll
\[-A+B+C =0\] \[B-2C=2 \] \[A+C =0 \]
UsukiDoll
  • 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
  • anonymous
B=2+2c A=2+2c+c A+c=0 2+2c+c+c=0
UsukiDoll
  • 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
  • anonymous
ok that is easier.
UsukiDoll
  • UsukiDoll
so now adding B-2C=2 B+2C=0 we have 2B=2 B = 1
UsukiDoll
  • 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
  • UsukiDoll
B-2C=2 1-2C=2 |dw:1438074079535:dw|
UsukiDoll
  • 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
  • anonymous
a=1/2
UsukiDoll
  • UsukiDoll
|dw:1438074167001:dw|
UsukiDoll
  • 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
  • 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
  • UsukiDoll
|dw:1438074628205:dw|
UsukiDoll
  • 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
  • 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
  • anonymous
Go on:)
anonymous
  • anonymous
yeah I also got those parts.
anonymous
  • anonymous
the last one takes more work.
UsukiDoll
  • UsukiDoll
yeah x.x now.. using u sub. u = x-1 du = 1 dx
UsukiDoll
  • UsukiDoll
|dw:1438074892358:dw| I wanna recheck that just to be sure
anonymous
  • anonymous
yep thats correct.
UsukiDoll
  • UsukiDoll
thanks :) so final answer after all that jazz is |dw:1438075034959:dw|
anonymous
  • anonymous
I will take a break and go through this all again. So Im 100 on this THANKS!!!!
UsukiDoll
  • UsukiDoll
you're welcome :)

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