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Math Help - 2 Integration problems

  1. #1
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    2 Integration problems

    Ughh confusing integrals ...

    1) Integrate x^2sqrt(x+2) dx

    2) Integrate x/(2-x^2) dx
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  2. #2
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    Quote Originally Posted by Hibijibi View Post
    Ughh confusing integrals ...

    1) Integrate x^2sqrt(x+2) dx
    With t^2 = x+2 substitution, we have 2t\,dt = \, dx

    \int x^2\sqrt{x+2} \,dx = \int 2t^2(t^2 - 2)^2 \, dt = \int (2t^6  - 8t^4+ 8t^2) \, dt

    Now it is no longer confusing, is it?
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  3. #3
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    Quote Originally Posted by Hibijibi View Post
    Ughh confusing integrals ...

    2) Integrate x/(2-x^2) dx
    Observe that d(x^2) = 2x \, dx

    So

    \int \frac{x}{2-x^2}\, dx = -\frac12 \int \frac{-2x}{2-x^2}\, dx = -\frac12 \int \frac1{2-x^2}\, d(2 - x^2)

    If this method looks weird to you, use the substitution u = 2 - x^2. I have done the same thing, but avoided introducing another variable
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  4. #4
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    2. \int \frac{x}{2-x^2}~dx

    Let u=2-x^2. Then du = -2x~dx.

    It becomes,

    \int -\frac{1}{2} \cdot \frac{1}{u}~du

    -\frac{1}{2}\ln |u|

    -\frac{1}{2}\ln |2-x^2|
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  5. #5
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    (2) Let {\color{blue}u = 2 - x^{2}} whose derivative is  du = -2x dx \: \Rightarrow \: {\color{red}x dx = -\frac{du}{2}}

    So the integral: \int \frac{{\color{red}xdx}}{{\color{blue}2-x^{2}}} = \int \frac{-\frac{du}{2}}{u} = -\frac{1}{2} \int \frac{du}{u}
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