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Thread: 2 Integration problems

  1. #1
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    2 Integration problems

    Ughh confusing integrals ...

    1) Integrate $\displaystyle x^2sqrt(x+2) dx$

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

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

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

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

    So

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

    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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    2. $\displaystyle \int \frac{x}{2-x^2}~dx$

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

    It becomes,

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

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

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

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