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Math Help - Help with integral

  1. #1
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    Help with integral

    How do I integrate (sqrt(x-2))(x2 + x + 1)

    I've got a feeling it's some basic method of integration that I'm forgetting, but I can't find anything like it in my notes.
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  2. #2
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    Re: Help with integral

    substitute :

    u=\sqrt{x-2}
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  3. #3
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    Re: Help with integral

    I tried that. It doesn't really get me anywhere.
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  4. #4
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    Re: Help with integral

    Quote Originally Posted by necromanzer52 View Post
    It doesn't really get me anywhere.
    It should. Post your working?
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  5. #5
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    Re: Help with integral

    u = rt(x-2)

    du/dx = (1/2)(x-2)^(1/2) = 1/(2(rt(x-2)))

    (2(rt(x-2))du = dx

    Subbing that back in gives me :

    u((x2 + x + 1)/((2(rt(x-2))) du

    I could cancel out the u to get (1/2)(x2 + x + 1) du, but that's no use.
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  6. #6
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    Re: Help with integral

    I would differentiate it in the form u^2=x-2 to get 2udu=dx

    You can also then rearrange to get x=u^2+2 then it all comes out nicely when you sub in.
    Thanks from necromanzer52
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  7. #7
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    Re: Help with integral

    Quote Originally Posted by necromanzer52 View Post
    How do I integrate (sqrt(x-2))(x2 + x + 1)

    I've got a feeling it's some basic method of integration that I'm forgetting, but I can't find anything like it in my notes.
    Start by writing

    \displaystyle \begin{align*} t &= x - 2 \\ x &= t + 2 \\ x^2 &= (t + 2)^2 \\ x^2 &= t^2 + 4t + 4 \\ x^2 + x &= t^2 + 4t + 4 + t + 2 \\ x^2 + x &= t^2 + 5t + 6 \\ x^2 + x + 1 &= t^2 + 5t + 7 \\ x^2 + x + 1 &= (x - 2)^2 + 5(x - 2) + 7  \end{align*}

    Now rewrite the integrand as \displaystyle \begin{align*} \frac{\sqrt{x - 2}}{x^2 + x + 1} = \frac{2(x - 2)}{2\sqrt{x - 2}\left[(x - 2)^2 + 5(x - 2) + 7\right] } \end{align*}

    Now if you make the substitution \displaystyle \begin{align*} u = \sqrt{x - 2} \implies du = \frac{1}{2\sqrt{x - 2}}\,dx \end{align*} the integral becomes

    \displaystyle \begin{align*} \int{\frac{2u^2}{u^4 + 5u^2 + 7}\,du} \end{align*}

    Can you go from here?
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  8. #8
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    Re: Help with integral

    I can go from there alright. I don't really follow you after you wrote "Now rewrite the integrand as". They're multiplying rather than dividing.

    But the other guys method works fine. So I'm grand.
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