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Thread: Integration by Substitution - Help

  1. #1
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    Integration by Substitution - Help

    Here is my original problem:

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

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

    $\displaystyle u = (1+x)$ ; $\displaystyle du=dx$ ; $\displaystyle x=u-1$

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

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

    Am I on the right track so far? Multiplying the $\displaystyle u^2$ with the $\displaystyle (u-1)^{\frac{-1}{2}}$ is kinda throwing me off...should that be what I do next? If so, as a result of that multiplication, I am getting $\displaystyle u^{\frac{3}{2}}-u^2$ 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 $\displaystyle u=\sqrt{x}$ instead?

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

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

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

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

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

    $\displaystyle \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$

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

    ..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 $\displaystyle \displaystyle\frac{du}{dx}$ concept is still hard for me to get my mind wrapped around. I don't think I really understand it.

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

    $\displaystyle \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$

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

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

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

    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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    $\displaystyle \displaystyle\int x^{\frac{3}{2}}+2x^{\frac{1}{2}}+x^{\frac{-1}{2}}dx=\frac{2}{5}x^{\frac{5}{2}}\frac{4}{3}x^{\ frac{3}{2}}+2x^{\frac{1}{2}}+c=2x^{\frac{1}{2}}(\f rac{1}{5}x^{2}+\frac{2}{3}x+1)+c$

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