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Partial fractions

  1. Sep 15, 2004 #1
    i will use "\int" as a integral sign since latex is down.

    \int (7)/(x^2-1)*dx

    using partial fractions...
    took out the 7...

    7\int (1)/(x+1)(x-1)

    A(x-1) + B(x+1) = 7

    if x = 1, B=7/2
    if x = -1, A= -7/2

    ok it's time to set up my integral function:

    7\int -7/2(x-1) + 7\int 7/2(x-1)

    take out the 7...

    14\int -1/2x-1 + 14\int 1/2x-2

    now to solve the integrals

    14(-ln(2x-2) + ln(2x-2)) + c

    where did i go wrong?
  2. jcsd
  3. Sep 15, 2004 #2
    See attachement.

    I would suggest spending 15 dollars (think this is the cost now) to get a passcode at www.calc101.com because it is a great learning tool. You will see what you are doing right or incorrect at anytime.

  4. Sep 15, 2004 #3


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    In one of your steps you mistakenly replaced x+1 with x-1.
  5. Sep 15, 2004 #4

    i thought it was pretty cool, but after looking at their answer, it was wrong lol. i submit my hw online, so it checks my answer.

    "In one of your steps you mistakenly replaced x+1 with x-1"

    "7\int -7/2(x-1) + 7\int 7/2(x-1)"

    ok changed that to

    7\int -7/2(x-1) + 7\int 7/2(x+1)

    take out the 7s
    14 \int -1/(2x-2) + 14\int 1/(2x+2)

    14 (-ln|2x-2| + ln |2x+2|)

    but it's wrong
  6. Sep 15, 2004 #5
    Hmm.... their answer looked correct to me. Looks like I need to do it the old fashion way .. by hand :) Remember, when doing a deriative or an integral you can always get more than one answer.
  7. Sep 15, 2004 #6
    yea, but at the end, it should be equal to each other.

    when i use maple, i get "-7*arctanh(x)"

    and it's correct(submitted online already), maple has always been the best tool, but it doesnt show you step by step on how it does it. it would be awesome if it did. but i would still like to know where i went wrong(i dont want it in the form arctanh, we havent really gone over them yet)
    Last edited: Sep 15, 2004
  8. Sep 16, 2004 #7


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    You also forgot to divide by two in one step. I suggest taking the derivative of your answer and comparing it with the integrand. I also recommend taking the derivative of the atanh(x) solution for comparison! :-)
  9. Sep 16, 2004 #8

    "7\int -7/2(x-1) + 7\int 7/2(x+1)"

    when i took out the 7's, i added it for some reason. ok now i just took out the 7/2

    49/2 (-ln(x-1) + ln(x+1))

    but its still the wrong answer.

    k i took the derivative of it, and know it's the wrong answer. but i cant really fix it cause i dont know what im doing wrong.
  10. Sep 16, 2004 #9
    calc101.com likes to take extra steps in factoring their answer for integrals, so the answer you may be getting won't look anything like what they have. Makes it tricky half the time to make sure it is the same.

    I paid the $15 for an account and have never had it give me a wrong answer. In fact it has helped me pick up on strategies to employ rather nicely, so if you are struggling I would strongly suggest dropping the cash for it, regardless of your impression from what you get for free.

    The only thing I don't like is that they'll use any equation possible to simplify, so while normally you might be doing some trig sub and integration by parts, every now and then you get an integral they just throw straight into a mold and pop out an answer. Still works, but doesn't help you figure it out as much as it suggest more formula memorization. Regardless, a large majority of the time it works very well.
  11. Sep 16, 2004 #10
    i got the answer now, i dont know what i was thinking setting A + B =7, i should of set it to 1, because i already took out the 7 in the beginning

    7/2(-ln|x-1| + ln|x+1|) is my answer... and it's correct, but cal101.com gets

    7/2*(-log(x+1) + log(x-1))

    they said that the integral of 1/t is log(t), but it's ln(t)!
    Last edited: Sep 16, 2004
  12. Sep 16, 2004 #11


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    Remember how you set up the original equation to determine A and B:
    Yhe last formula is wrong. First, because the coefficient "7" is included already in A and B. Second: you got the equation for A, B by replacing the original integrand by

    7/[(x-1)(x+1)] = A/(x+1) + B/(x-1) ---> 7=A(x-1)+B(x+1)

    So your integral becomes:

    S [-(7/2)/(x+1)] dx+ S [(7/2)/(x-1)] dx

    Factor out 7/2:

    (7/2)S [-1/(x+1)] dx+ S [1/(x-1)] dx = (7/2)[-ln|x+1|+ln|x-1|]+c

    The derivative of the result is (7/2)*[-l/(x+1)+1/(x-1)]=
    =(7/2)*[-(x-1)+(x+1)]/(x^2-1)=(7/2)*(-x+1+x+1)/(x^2-1) = (7/2)*2/(x^2-1) and this is the original function.

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