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Thread: 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?

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

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

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

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

    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 $\displaystyle t = \sqrt{x}$ so that $\displaystyle dx = 2\sqrt{x} \;{dt}$ you will end up with:

    $\displaystyle \displaystyle \begin{aligned} I & = 2\int\frac{1}{t^2-1}\;{dt} = \int\frac{(t+1)-(t-1)}{(t-1)(t+1)}\;{dt} = \int\frac{1}{t-1}-\frac{1}{t+1}\;{dt} \\& = \ln\left(t-1\right)-\ln\left(t+1\right)+k = \ln\left|\frac{t-1}{t+1}\right|+k = \ln\left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|+k. \end{aligned}$
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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 $\displaystyle \displaystyle x^3$...
    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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