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Math Help - Integration by Substitution - Help

  1. #1
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    Integration by Substitution - Help

    Here is my original problem:

    \displaystyle\int\frac{(1+x)^2}{\sqrt{x}} dx

    Here are the steps I worked. I'm VERY new to integrals and substitution, so bear with me, please.

    u = (1+x) ; du=dx ; x=u-1

    \displaystyle\int u^2(x)^{\frac{-1}{2}}du

    \displaystyle\int u^2(u-1)^{\frac{-1}{2}}du

    Am I on the right track so far? Multiplying the u^2 with the (u-1)^{\frac{-1}{2}} is kinda throwing me off...should that be what I do next? If so, as a result of that multiplication, I am getting u^{\frac{3}{2}}-u^2 which I think is wrong. I don't mean this to be rude but if you want to help me, please dont work the problem for me, I just want to be pointed in the correct direction. I learn better when I can work it out myself. Thanks
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  2. #2
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    Why don't you try the substitution u=\sqrt{x} instead?

    \displaystyle\int u^2(u-1)^{\frac{-1}{2}}du is no easier to integrate than \displaystyle\int\frac{(1+x)^2}{\sqrt{x}} dx.
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  3. #3
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    ok. so if I do that, then I end up with:

    \displaystyle\int\frac{(1+x)^2}{u}du=\int(1+x)^2*u  ^{-1}du

    Where would I go from here? I can't integrate the terms seperately.
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  4. #4
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    Remember that if \displaystyle u = \sqrt{x} then \displaystyle x = u^2.

    You also need to recognise where \displaystyle \frac{du}{dx} is in your function.
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  5. #5
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    that whole \displaystyle\frac{du}{dx} concept is still hard for me to get my mind wrapped around. I don't think I really understand it.

    Since \displaystyle x = u^2 and \displaystyle u = \sqrt{x} then...

    \displaystyle\int\(1+x)^2*u^{-1}du=\int(x^2+2x+1)u^{-1}du=\int(u^4+2u^2+1)u^{-1}du

    \displaystyle=\int(u^3+2u+u^{-1})du=

    ..is this right so far?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dbakeg00 View Post
    that whole \displaystyle\frac{du}{dx} concept is still hard for me to get my mind wrapped around. I don't think I really understand it.

    Since \displaystyle x = u^2 and \displaystyle u = \sqrt{x} then...

    \displaystyle\int\(1+x)^2*u^{-1}du=\int(x^2+2x+1)u^{-1}du=\int(u^4+2u^2+1)u^{-1}du

    \displaystyle=\int(u^3+2u+u^{-1})du=

    ..is this right so far?
    you could have used this same trick from the very beginning. a substitution wasn't necessary.

    \displaystyle \frac {(1 + x)^2}{\sqrt x} = \frac {1 + 2x + x^2}{\sqrt x} = \frac 1{\sqrt x} + \frac {2x}{\sqrt x} + \frac {x^2}{\sqrt x} = x^{-1/2} + 2x^{1/2} + x^{3/2}

    now just integrate the last form using the good ol' fashioned power rule for integrals
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  7. #7
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    \displaystyle\int x^{\frac{3}{2}}+2x^{\frac{1}{2}}+x^{\frac{-1}{2}}dx=\frac{2}{5}x^{\frac{5}{2}}\frac{4}{3}x^{\  frac{3}{2}}+2x^{\frac{1}{2}}+c=2x^{\frac{1}{2}}(\f  rac{1}{5}x^{2}+\frac{2}{3}x+1)+c

    Is that correct?
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  8. #8
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    Yes, that is correct.
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