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Math Help - Integration Help

  1. #1
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    Integration Help

    I need to integrate 1/(16x^3-4x^2+4x-1)

    I first split the denominator into (4x-1)(4x^2+1) and used partial fraction decomposition to get:

    the integral of (4/5)/(4x-1)+((-4/5)x-(1/5))/(4x^2+1)

    I know how to solve the first segment of the integral, but I don't know how to integrate ((-4/5)x-(1/5))/(4x^2+1)
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  2. #2
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    Quote Originally Posted by Simon777 View Post
    I need to integrate 1/(16x^3-4x^2+4x-1)

    I first split the denominator into (4x-1)(4x^2+1) and used partial fraction decomposition to get:

    the integral of (4/5)/(4x-1)+((-4/5)x-(1/5))/(4x^2+1)

    I know how to solve the first segment of the integral, but I don't know how to integrate ((-4/5)x-(1/5))/(4x^2+1)
    Try splitting it into two rational expressions - the first with (-4/5)x in the numerator and the second with -(1/5) in the numerator.
    The first one will involve ln(4x^2+1),
    the second will need substitution.
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  3. #3
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    So you have:

    \int \frac{4}{5(4x-1)}-\frac{4x+1}{5(x^{2}+1)}dx \Rightarrow

    \int \frac{4}{5(4x-1)}-\frac{4x}{5(x^{2}+1)}+\frac{1}{5(x^{2}+1)}dx

     u = 4x-1; du = 4dx

    w = x^2+1; dw = 2xdx

    \int \frac{du}{5u}-\frac{2dw}{5w}+\frac{1}{5(x^{2}+1)} \Rightarrow

    \frac{ln|4x-1|-ln(x^2+1)^{2}+tan^{-1}(x)}{5}
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  4. #4
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    Quote Originally Posted by ANDS! View Post
    So you have:

    \int \frac{4}{5(4x-1)}-\frac{4x+1}{5(x^{2}+1)}dx \Rightarrow
    With everything after the subtraction sign, why is it just (x^2+1)? Shouldn't it be (4x^2+1)? That was what the original problem was split into.
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  5. #5
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    Quote Originally Posted by Debsta View Post
    Try splitting it into two rational expressions - the first with (-4/5)x in the numerator and the second with -(1/5) in the numerator.
    The first one will involve ln(4x^2+1),
    the second will need substitution.
    Thank you, I got a little farther with that advise, but what is the integral of:

    1/(4x^2+1)

    I can't do u substitution and I can't convert this into tan^-1 because of the 4.
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  6. #6
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    Quote Originally Posted by Simon777 View Post
    Thank you, I got a little farther with that advise, but what is the integral of:

    1/(4x^2+1)

    I can't do u substitution and I can't convert this into tan^-1 because of the 4.
    edit: mistake. refer to skeeter's post
    Last edited by Defunkt; March 27th 2010 at 09:36 AM.
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  7. #7
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    Quote Originally Posted by Simon777 View Post
    Thank you, I got a little farther with that advise, but what is the integral of:

    1/(4x^2+1)

    I can't do u substitution and I can't convert this into tan^-1 because of the 4.
    \frac{1}{4x^2+1} = \frac{1}{(2x)^2+1} = \frac{1}{2} \cdot \frac{2}{(2x)^2+1}

    \int \frac{1}{2} \cdot \frac{2}{(2x)^2+1} \, dx

    \frac{1}{2}\arctan(2x) + C
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  8. #8
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    Quote Originally Posted by Simon777 View Post
    With everything after the subtraction sign, why is it just (x^2+1)? Shouldn't it be (4x^2+1)? That was what the original problem was split into.
    It is. Doesn't change using trig sub though. Even if it wasn't a perfect square.
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  9. #9
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    Thank you all so much for the help. Can someone please check my answer?

    ln|((4x-1)^(1/5)) / ((4x^2+1)^(1/10))| - (1/10)arctan(2x) + C
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    Quote Originally Posted by Simon777 View Post
    Thank you all so much for the help. Can someone please check my answer?

    ln|((4x-1)^(1/5)) / ((4x^2+1)^(1/10))| - (1/10)arctan(2x) + C
    Yes that looks right. Just one thing, because of the yukky exponents in the first term, I would have written the first term as:
    (1/5) ln|(4x-1)/sqr(4x^2+1)|
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  11. #11
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    Quote Originally Posted by Debsta View Post
    Yes that looks right. Just one thing, because of the yukky exponents in the first term, I would have written the first term as:
    (1/5) ln|(4x-1)/sqr(4x^2+1)|
    Thanks for taking the time to check that.
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