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anonymous
 one year ago
Express following in partial fractions...evaluate the given integral
anonymous
 one year ago
Express following in partial fractions...evaluate the given integral

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438068450403:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3oh crap... let me expand (x1)^2(x+1) first with wolfram and see if it comes back as factorable all the way.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3ok 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
 one year ago
Best ResponseYou've already chosen the best response.3for example dw:1438068733033:dw then take lcd of all of these guys

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3luckily the problem we are given isn't harsh like that (though I did s^5 at one point)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3show me your attempt :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2x }{ (x2)^2(x+1) }=\frac{ A }{ x+1 }+\frac{ B }{(x2)^2 }\] simplify \[2x=\frac{ A(x+2 )((x2)^2) }{ x+2 } + \frac{ B((x2)^2(x+2)) }{ (x2)^2 }\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3the A part is fine.. the one with the exponent.. no...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[2x=A(x2)^2+B(x+1)\] x=2 gives \[4=A(0)^2+B(3)\] B=4/3

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3whenever 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
 one year ago
Best ResponseYou've already chosen the best response.3I 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
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438069450631:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I think order matters too... dw:1438069483088:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3the lcd for this is \[(x+1)(x1)^2 \] alright take the lcd ;)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3ahem... take the lcd for dw:1438069618368:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0alright alright, give me a sec:)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438069792243:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3stop setting 2x = that's not going to work

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3the lcd part is good though

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438069902990:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3k expand all of that

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3*eats popcorn while I wait*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont get this at all....whats wrong with me!!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438070102575:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whats does foil mean':)?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3foil  first outer inner last... okkk expand :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Maybe I shouldve told u havent slept in 2 days...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3then you should take a break from math and sleep

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its 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
 one year ago
Best ResponseYou've already chosen the best response.3try finishing the problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so expanding c gives x^22x+1

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438070573305:dw dw:1438070618427:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438070677726:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3now distribute the letters

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im 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
 one year ago
Best ResponseYou've already chosen the best response.3the problem is that I can't give direct answers... so can't finish it without getting busted.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3@Xlegalize can you come back over here so we can finish this please?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3ok so the last step we left off is dw:1438073386251:dw so what we need to do next is to distribute the letters

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I'll give an example \[A(x^21) = Ax^2A \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3then similarly for B and C \[B(x+1) \rightarrow Bx+B \] \[C(x^22x+1) \rightarrow Cx^2C2x+C \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3yeahhh that 2xC combination looks gross... and awkward XD anyway now we can put that long string in the numerator .

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[\large \frac{Ax^2A+Bx+B+Cx^22xC+C}{(x+1)(x1)^2}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw: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
 one year ago
Best ResponseYou've already chosen the best response.3\[A+B+C =0\] \[B2C=2 \] \[A+C =0 \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so 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
 one year ago
Best ResponseYou've already chosen the best response.0B=2+2c A=2+2c+c A+c=0 2+2c+c+c=0

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3let'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

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so now adding B2C=2 B+2C=0 we have 2B=2 B = 1

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so 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
 one year ago
Best ResponseYou've already chosen the best response.3B2C=2 12C=2 dw:1438074079535:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so 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

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438074167001:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3that'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
 one year ago
Best ResponseYou've already chosen the best response.3\[A+C =0 \] since A = 1/2 and C = 1/2 \[\frac{1}{2}\frac{1}{2}=0 \] \[0=0 \] \[A+B+C =0 \] \[\frac{1}{2}+1\frac{1}{2} =0\] \[\frac{1}{2}\frac{1}{2}+1 =0\] \[\frac{1}{2}\frac{1}{2}+1 =0\] \[\frac{2}{2}+1 =0\] \[1+1 =0\] \[0 =0\] \[B2C=2 \] B = 1 \[12\frac{1}{2}=2 \] \[1+\frac{2}{2}=2 \] \[1+1=2 \] \[2=2 \] ok after that long verification latex string we can plug our values into the earlier equation and integrate

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438074628205:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so after all that lcd solving for a b c we can finally integrate... YAY! 2 of them are logs.. one is u substitution

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438074757028:dw so the red portion is the \[\frac{1}{(x1)^2}\] part. we have to do a bit more work on it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah I also got those parts.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the last one takes more work.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3yeah x.x now.. using u sub. u = x1 du = 1 dx

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438074892358:dw I wanna recheck that just to be sure

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3thanks :) so final answer after all that jazz is dw:1438075034959:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I will take a break and go through this all again. So Im 100 on this THANKS!!!!
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