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Math Help - Integration

  1. #1
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    Integration

    ∫ (3x-1)/(x^2-2x+10)^0.5 dx


    How can I integrate this?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    \frac{3x-1}{\sqrt{x^2-2x+10}}=\frac{3x-1}{\sqrt{x^2-2x+1+9}}=(3x-1)\cdot\frac{1}{\sqrt{(x-1)^2+9}}

    Transforming the fraction to make appear \mathrm{arcsinh}'u=\frac{1}{\sqrt{u^2+1}} may lead to the result. Any idea on how it can be achieved ?
    Last edited by flyingsquirrel; May 17th 2008 at 02:23 AM. Reason: arccosh -> arcsinh
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  3. #3
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    Quote Originally Posted by flyingsquirrel View Post
    Any idea on how it can be achieved ?
    Actually, no. Can you tell me, please?
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  4. #4
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    \frac{3x-1}{\sqrt{x^2-2x+10}}=\frac{3x-1}{\sqrt{x^2-2x+1+9}}=(3x-1)\cdot\frac{1}{\sqrt{(x-1)^2+9}}

    Transforming the fraction to make appear \mathrm{arccosh}'u=\frac{1}{\sqrt{u^2+1}} may lead to the result. Any idea on how it can be achieved ?
    I know that, 1/√(1+x^2) = arsinh x.

    But I donít know what is the next step of (3x-1)\cdot\frac{1}{\sqrt{(x-1)^2+9}}
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by geton View Post
    I know that, 1/√(1+x^2) = arsinh x.

    But I don’t know what is the next step of (3x-1)\cdot\frac{1}{\sqrt{(x-1)^2+9}}
    We want the fraction to be \frac{1}{\sqrt{u^2+1}} and the square is already present. To make appear 1, let's factor by 9 inside the square root :

    (3x-1)\cdot\frac{1}{\sqrt{(x-1)^2+9}}=(3x-1)\cdot\frac{1}{\sqrt{9\left(\frac{(x-1)^2}{9}+1\right)}}=(3x-1)\cdot\frac{1}{3\sqrt{\left(\frac{x-1}{3}\right)^2+1}}

    Hence \frac{3x-1}{\sqrt{x^2-2x+10}}=(3x-1)\cdot\frac{1}{3\sqrt{\left(\frac{x-1}{3}\right)^2+1}}

    Substituting u=\frac{x-1}{3} makes appear \mathrm{arcsinh}'u... but it is multiplied by a polynomial. How can one get rid off it ?
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  6. #6
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    Quote Originally Posted by geton View Post
    ∫ (3x-1)/(x^2-2x+10)^0.5 dx


    How can I integrate this?
     \frac{3x - 1}{\sqrt{x^2 - 2x + 10}} = \frac{3 \left( x - \frac{1}{3} \right)}{\sqrt{x^2 - 2x + 10}} = \frac{\frac{3}{2} \left( 2x - \frac{2}{3} \right)}{\sqrt{x^2 - 2x + 10}}


    = \frac{\frac{3}{2} \left( [2x - 2] + 2 - \frac{2}{3} \right)}{\sqrt{x^2 - 2x + 10}} = \frac{3}{2} \, \frac{2x - 2}{\sqrt{x^2 - 2x + 10}} + \frac{3}{2} \, \frac{2 - \frac{2}{3}}{\sqrt{x^2 - 2x + 10}}


    = \frac{3}{2} \, \frac{2x - 2}{\sqrt{x^2 - 2x + 10}} + \frac{2}{\sqrt{x^2 - 2x + 10}}.


    Integral of the first term is done using an obvious substitution. Integral of the second term is done by completing the square inside the square root etc.
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  7. #7
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    Thank you so much mr fantastic for help. And sorry for late reply, because of study pressure.
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