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Math Help - sustitution intregration again

  1. #1
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    sustitution intregration again

    I think I am getting closer, but necessarily correct.

    integral (10x^5-5x)/square root (x^4-x^2+6) dx

    let x = x^4 -x^2 +6
    d/dx = 4x^3 - 2x or 2x(2x^2 -1)

    1/2 integral x^-1/2 du
    1/2 *(u^1/2)/1/2 + C
    1/4*u^1/2 + C
    1/4 (x^4-x^2+6)^1/2 +C
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  2. #2
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    Quote Originally Posted by startingover View Post
    I think I am getting closer, but necessarily correct.

    integral (10x^5-5x)/square root (x^4-x^2+6) dx
    Check you integral again. I think you made a mistake on it.
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  3. #3
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    THP:
    Man, you are good....you are right, I didn't write down the problem correctly.

    integral (10x^3-5x)/square root (x^4-x^2+6) dx
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  4. #4
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    To find I = \int \frac{10x^3-5x}{\sqrt{x^4-x^2+6}} dx = \frac{5}{2}\int \frac{4x^3-2x}{\sqrt{x^4-x^2+6}} dx:


    Substitute u = x^4 - x^2 + 6~~~\Rightarrow~~~du = (4x^3 - 2x) dx.

    Then I = \frac{5}{2} \int \frac{1}{\sqrt{u}} du = \frac{5}{4}\sqrt{u} + C = \frac{5}{4}\sqrt{x^4 - x^2 + 6} + C.
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  5. #5
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    The most confusion I face is in figuring out how you get the 5/2...
    I understand that (x^4-X^2+6)^-1/2
    Then it becomes 1/n+1^n+1
    I assume it is because the numerator becomes 5x(2x^2-1)...
    Please advise.
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  6. #6
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    Quote Originally Posted by startingover View Post
    The most confusion I face is in figuring out how you get the 5/2...
    I understand that (x^4-X^2+6)^-1/2
    Then it becomes 1/n+1^n+1
    I assume it is because the numerator becomes 5x(2x^2-1)...
    Please advise.
    OK. Right at the beginning. I decided to "bring outside" a factor of 5/2 on the first line. Like this:

    \int \frac{10x^3-5x}{\sqrt{x^4-x^2+6}} dx ~~=~~ \frac{5}{2} \cdot \int \frac{(\frac{2}{5})\left(10x^3-5x\right)}{\sqrt{x^4-x^2+6}} dx ~~=~~ \frac{5}{2} \cdot \int \frac{(4x^3-2x)}{\sqrt{x^4-x^2+6}} dx

    Hopefully, from that line above, you can see how I did it. The more important question, I guess, is why I did it!

    The reason is, because then it's easy to see the substitution you should use.

    After you "bring out" the 5/2, the remaining integral has a particular, special form:

     \int \frac{(4x^3-2x)}{\sqrt{x^4-x^2+6}} dx ~~=~~ \int f(u) \left( \frac{du}{dx} \right) dx

    Here, u = x^4 - x^2 + 6;

    f(u) = \frac{1}{\sqrt{u}} ~=~ \frac{1}{\sqrt{x^4 - x^2 + 6}};

    \left( \frac{du}{dx}\right) = (4x^3 - 2x).

    This whole process of switching (or substituting) u for x is called 'integration by substitution'.

    Then, it becomes easy to finish the integral, because you can say:

    \int f(u) \left( \frac{du}{dx} \right) dx ~~=~~ \int f(u) du

    and because you know f(u) = \frac{1}{\sqrt{u}} = u^{-1/2}, you can integrate it using the normal rule:

     \int u^n du ~~=~~ \frac{u^{n+1}}{n+1} + C.

    - - -

    I hope that helps to explain what I was doing. Still, if there's a specific bit you don't understand, I'll "zoom in" on it and explain further.
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  7. #7
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    \int {\frac{{10x^3 - 5x}}<br />
{{\sqrt {x^4 - x^2 + 6} }}~dx} = 5\int {\frac{{2x^3 - x}}<br />
{{\sqrt {x^4 - x^2 + 6} }}~dx} = \frac{5}<br />
{2}\int {\frac{{4x^3 - 2x}}<br />
{{\sqrt {x^4 - x^2 + 6} }}~dx}


    It's faster if you only take u=\sqrt{x^4 - x^2 + 6}
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  8. #8
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     I = \frac{5}{2} \int \frac{1}{\sqrt{u}}du
     I = \frac{5}{2} . \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}

     I = \frac{5}{2} . \frac{\sqrt{u}}{\frac{1}{2}}


     I = 5 . \sqrt{u}

     I = 5 . \sqrt{x^4-x^2+6}
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  9. #9
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post

     I = 5 . \sqrt{x^4-x^2+6}
    You are right, I was wrong by that factor of 4.
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