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Math Help - Integration

  1. #1
    Senior Member slevvio's Avatar
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    Integration

    Hello, I was wondering if I could get some help with this integral - it would be much appreciated - thanks.

    \int{\frac{1}{\sqrt{x^2-4x+13}}}dx = \int{\frac{1}{\sqrt{(x-2)^2+9}}}dx
    Last edited by slevvio; December 27th 2007 at 08:44 AM.
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  2. #2
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by slevvio View Post
    Hello, I was wondering if I could get some help with this integral - it would be much appreciated - thanks.

    \int{\frac{1}{\sqrt{x^2-4x+13}}}dx = \int{\frac{1}{(x-2)^2+9}}dx
    Do you mean this?

    \int{\frac{1}{\sqrt{x^2-4x+13}}}dx = \int{\frac{1}{\sqrt{(x-2)^2+9}}}dx

    If so, then you can use a trig substitution for x - 2. Set it equal to 3*tan(u). Everything else should fall into place for easy integration!

    EDIT: Soroban posted shortly after I did, check out his detailed solution!
    Last edited by colby2152; December 26th 2007 at 07:05 AM.
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  3. #3
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    Hello, slevvio!

    \int \frac{dx}{\sqrt{x^2-4x+13}} \:= \:\int \frac{dx}{\sqrt{(x-2)^2+9}}

    Let: . x-2 \:=\:3\tan\theta\quad\Rightarrow\quad dx \:=\:3\sec^2\!\theta\,d\theta

    . . and: . \sqrt{(x-2)^2+9} \;=\;\sqrt{9\tan^2\!\theta  + 9} \:=\:\sqrt{9(\tan^2\!\theta + 1)} \:=\:\sqrt{9\sec^2\!\theta} \:=\:3\sec\theta


    Substitute: . \int\frac{3\sec^2\!\theta\,d\theta}{3\sec\theta} \;=\;\int\sec\theta\,d\theta \;=\;\ln|\sec\theta + \tan\theta| + C

    . . then back-substitute . . .

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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    We have a hidden natural logarithm here.

    Quote Originally Posted by slevvio View Post
    \int{\frac{1}{\sqrt{x^2-4x+13}}}dx
    \int {\frac{{dx}}<br />
{{\sqrt {x^2  - 4x + 13} }}}  = \int {\frac{{dx}}<br />
{{\sqrt {(x - 2)^2  + 9} }}} .

    Substitute u = x + \sqrt {(x - 2)^2  + 9}  - 2 \implies du = 1 + \frac{{x - 2}}<br />
{{\sqrt {(x - 2)^2  + 9} }}\,dx.

    So \frac{{dx}}<br />
{{\sqrt {(x - 2)^2  + 9} }} = \frac{{du}}<br />
{u}.

    The rest follows.
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