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Math Help - indefinite integral

  1. #1
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    indefinite integral

    Hi, I've attached an indefinite integral problem I thought I did correctly, but the answer turned out to be wrong. If anyone could point out my error I would greatly appreciate it! Thanks

    indefinite integral-intprob.jpg
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    Re: indefinite integral

    I would do it a different way...

    \displaystyle \begin{align*} \int{\frac{\sqrt{x-1}}{x}\,dx} &= \int{\frac{(\sqrt{x-1})^2}{x\sqrt{x - 1}}\,dx} \\ &= 2\int{\frac{x - 1}{x}\cdot \frac{1}{2\sqrt{x - 1}}\,dx} \end{align*}

    Now make the substitution \displaystyle u = \sqrt{x - 1} \implies du = \frac{1}{2\sqrt{x - 1}}\,dx and also \displaystyle x = u^2 + 1, then the integral becomes

    \displaystyle \begin{align*} 2\int{\frac{x - 1}{x} \cdot \frac{1}{2\sqrt{x - 1}}\,dx} &= 2\int{\frac{u^2}{u^2 + 1}\,du} \\ &= 2\int{1 - \frac{1}{u^2 + 1}\,du} \\ &= 2\left(u - \arctan{u}\right) + C \\ &= 2\left[\sqrt{x-1} - \arctan{\left(\sqrt{x-1}\right)}\right] + C \end{align*}
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    Re: indefinite integral

    Your first substitution is not correct, you just said: u=sqrt(1-x) and substitued without calculating correctly du in function of dx.

    EDIT:
    Solution with your substitution(s):
    \int \frac{\sqrt{x-1}}{x}dx=\int \frac{\sqrt{x-1}}{(x-1)+1}dx
    Let \sqrt{x-1}=t so \frac{dx}{2\sqrt{x-1}} \Leftrightarrow dx=2tdt
    So the integral becomes:
    \int \frac{2t^2dt}{t^2+1}dt=2\int \frac{(t^2+1)-1}{t^2+1}dt=2\int \frac{t^2+1}{t^2+1}dt - 2\int \frac{dt}{t^2+1}=2t-2\arctan(t) + C = 2\sqrt{x-1}-2\arctan\left(\sqrt{x-1}\right)+C
    Last edited by Siron; July 13th 2011 at 07:48 AM.
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    Re: indefinite integral

    Write \int{\frac{\sqrt{x-1}}{x}\,dx}=\int{\frac{x-1}{x\sqrt{x-1}}\,dx}=\int{\frac{dx}{\sqrt{x-1}}}-\int{\frac{dx}{x\sqrt{x-1}}}, in the last integral put x=\frac1{t^2}, so the original integral equals 2\left( \sqrt{x-1}+\arcsin\frac1{\sqrt x} \right)+k.
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    Re: indefinite integral

    Quote Originally Posted by Krizalid View Post
    Write \int{\frac{\sqrt{x-1}}{x}\,dx}=\int{\frac{x-1}{x\sqrt{x-1}}\,dx}=\int{\frac{dx}{\sqrt{x-1}}}-\int{\frac{dx}{x\sqrt{x-1}}}, in the last integral put x=\frac1{t^2}, so the original integral equals 2\left( \sqrt{x-1}+\arcsin\frac1{\sqrt x} \right)+k.
    I think you've made a mistake, if you let \displaystyle x = t^{-2} then \displaystyle dx = -2t^{-3}\,dt, or \displaystyle dt = -\frac{1}{2}t^3\,dx... I'm not sure how you can convert \displaystyle \int{\frac{dx}{x\sqrt{x-1}}} to \displaystyle \int{\frac{dt}{\sqrt{t^2 - 1}}}...
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    Re: indefinite integral

    Try it again.
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    Re: indefinite integral

    Quote Originally Posted by Krizalid View Post
    Try it again.
    Well I realise now that if \displaystyle x = t^{-2} and \displaystyle dt = -\frac{1}{2}t^3\,dx then \displaystyle dt = -\frac{1}{2}x^6\,dx... Still don't see how you're going to use this substitution to your advantage...
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    Re: indefinite integral

    You don't see it?

    \int{\frac{dx}{x\sqrt{x-1}}}=-2\int{\frac{1}{{{t}^{3}}}\cdot \frac{1}{\frac{1}{{{t}^{2}}}\sqrt{\frac{1}{{{t}^{2  }}}-1}}\,dt}=-2\int{\frac{{{t}^{3}}}{{{t}^{3}}\sqrt{1-{{t}^{2}}}}\,dt}=-2\int{\frac{dt}{\sqrt{1-{{t}^{2}}}}}.
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    Re: indefinite integral

    Quote Originally Posted by Krizalid View Post
    You don't see it?

    \int{\frac{dx}{x\sqrt{x-1}}}=-2\int{\frac{1}{{{t}^{3}}}\cdot \frac{1}{\frac{1}{{{t}^{2}}}\sqrt{\frac{1}{{{t}^{2  }}}-1}}\,dt}=-2\int{\frac{{{t}^{3}}}{{{t}^{3}}\sqrt{1-{{t}^{2}}}}\,dt}=-2\int{\frac{dt}{\sqrt{1-{{t}^{2}}}}}.
    OK I see it now, thanks...
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  10. #10
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    Re: indefinite integral

    We coulda done the remaining integral by putting t=\sqrt{x-1}, which gives an arctangent.
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    Re: indefinite integral

    Quote Originally Posted by Krizalid View Post
    We coulda done the remaining integral by putting t=\sqrt{x-1}, which gives an arctangent.
    Which I did...
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  12. #12
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    Re: indefinite integral

    But on my remaining integral it leads a direct integral, in your remaining integral you had to do a long division, so it's one less step.
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