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Math Help - Partial fraction integration help!

  1. #1
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    Partial fraction integration help!

    Alright, these have been giving me heck all day. I have two short ones left. It's been a long day full of beating up on these monstrous ogres of problems ( Something like a 6x5 matrix for one of them ) and I'm glad to say these should be awesomely simple.

    First, The integral of 1/(x(x^4+1) dx. My book has a hint: "Multiply by x^3/x^3" on this problem. I broke it out into A/X + (B*X+C)/(x^4+1) = 1/(x(x^4+1)). Then I eliminated fractions. It's pretty easy to see A=1.

    Then I was promptly lost in the woods.

    Second one that looks a bit more intimidating is the integral of 1/(x^(3/2) - x^(1/2)).

    I started by letting U=x^(1/2) and DU=1/2(x^(1/2)) I ran through this one to the end, and ended up with LN(|U-1|/|u|)+C but that denominator should have been |u+1|.


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  2. #2
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    Why didn't you do what was suggested?

    \displaystyle \int{\frac{1}{x(x^4 + 1)}\,dx} = \int{\frac{x^3}{x^4(x^4 + 1)}\,dx}.

    Now make a very simple substitution...
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    Simply let t = \frac{1}{x}, then your integral:

    \begin{aligned}&  I  = -\int\frac{1/t^2}{1/t\left(\frac{1}{t^4}+1}\right)}\;{dt} = -\int\frac{t^3}{1+t^4}\;{dt} = -\frac{1}{4}\int\frac{4t^3}{1+t^3}\;{dt} \\ & = -\frac{1}{4}\int\frac{(1+t^4)'}{1+t^4}\;{dt} = -\frac{1}{4}\ln\left(1+t^4\right)+k = -\frac{1}{4}\ln\left(1+\frac{1}{x^4}\right)+k. \end{aligned}
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    Quote Originally Posted by TheCoffeeMachine View Post
    Simply let t = \frac{1}{x}, then your integral:

    \begin{aligned}&  I  = -\int\frac{1/t^2}{1/t\left(\frac{1}{t^4}+1}\right)}\;{dt} = -\int\frac{t^3}{1+t^4}\;{dt} = -\frac{1}{4}\int\frac{4t^3}{1+t^3}\;{dt} \\ & = -\frac{1}{4}\int\frac{(1+t^4)'}{1+t^4}\;{dt} = -\frac{1}{4}\ln\left(1+t^4\right)+k = -\frac{1}{4}\ln\left(1+\frac{1}{x^4}\right)+k. \end{aligned}
    Of course, this goes against what the OP was told to do in the problem, i.e. to multiply top and bottom by \displaystyle x^3...

    The substitution made after doing this is also much easier...
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    ROFL.

    Okay, I assumed they meant "Multiply after doing something to it.".

    I think my problem there may be of the T=pi/2 AM ID(10T) variety.

    I knew it was something easy. I'm almost afraid to ask about the other problem but I do need to practice!
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    Quote Originally Posted by Wolvenmoon View Post
    I knew it was something easy. I'm almost afraid to ask about the other problem but I do need to practice!
    Let t = \sqrt{x} so that dx = 2\sqrt{x} \;{dt} you will end up with:

    \displaystyle \begin{aligned} I & = 2\int\frac{1}{t^2-1}\;{dt} = \int\frac{(t+1)-(t-1)}{(t-1)(t+1)}\;{dt} = \int\frac{1}{t-1}-\frac{1}{t+1}\;{dt} \\& = \ln\left(t-1\right)-\ln\left(t+1\right)+k = \ln\left|\frac{t-1}{t+1}\right|+k = \ln\left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|+k. \end{aligned}
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    Quote Originally Posted by Prove It View Post
    Of course, this goes against what the OP was told to do in the problem, i.e. to multiply top and bottom by \displaystyle x^3...
    Of course, I was just pointing out another way that it could be done.
    The substitution made after doing this is also much easier.
    To be fair, I think both substitutions are very easy.
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