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Math Help - Anti derivative

  1. #1
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    Anti derivative

    I have to find the anti derivative of x(x^2+2)^2
    This is what I did:

    f ' (x) = ( x^2 + 2)^2

    = X (X^4 + 4x^2 + 4)

    F(x) = (1/2x^2) (1/5x^5 + 4/3x^3 + C)

    The book did not show any examples of problems like this and I was not sure if there was a product rule that I needed or if the Chain Rule came into play somehow.

    Thanks.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PensFan10 View Post
    I have to find the anti derivative of x(x^2+2)^2
    This is what I did:

    f ' (x) = ( x^2 + 2)^2

    = X (X^4 + 4x^2 + 4)

    F(x) = (1/2x^2) (1/5x^5 + 4/3x^3 + C)

    The book did not show any examples of problems like this and I was not sure if there was a product rule that I needed or if the Chain Rule came into play somehow.

    Thanks.
    \int{x(x^2+2)^2{dx}}

    Way one

    x(x^2+2)^2=x(x^4+4x^2+4)=x^5+4x^3+4x

    From there I am sure you can integrate

    Or you can do this

    Let u=x^2+2

    so du=2x\text{ }dx

    So we have

    \frac{1}{2}\int{u^2}du=\frac{u^3}{6}+C

    Now back sub.
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  3. #3
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    okay cool

    Okay so the mistake I made was that I should have distributed the x. I appreciate the speedy and helpful response.
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  4. #4
    Moo
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    Hello

    Quote Originally Posted by PensFan10 View Post
    I have to find the anti derivative of x(x^2+2)^2
    This is what I did:

    f ' (x) = x( x^2 + 2)^2

    = X (X^4 + 4x^2 + 4)

    F(x) = (1/2x^2) (1/5x^5 + 4/3x^3 + C) << SiMoon says : the antiderivative of a product is not the product of the antiderivatives ! If you wanted to do it this way, expan x(x^4+4x^2+4)=x^5+4x^3+4x and then integrate the sum.

    The book did not show any examples of problems like this and I was not sure if there was a product rule that I needed or if the Chain Rule came into play somehow.

    Thanks.
    Let f(x)=x^2+2.

    You can notice that f'(x)=2x.

    So x=\frac{f'(x)}{2}

    You have to find the antiderivative of x(x^2+2)=\frac 12 \cdot f'(x) \cdot (f(x))^2

    Note that the derivative of [f(x)]^n=n \cdot f'(x) \cdot [f(x)]^{n-1}

    Here, n-1=2. So n=3.

    --->< \frac 12 \cdot f'(x) \cdot [f(x)]^2 is the derivative of \frac 12 \cdot \frac 13 \cdot [f(x)]^3


    Antiderivative of x(x^2+2)^2 is \boxed{\frac 16 \cdot (x^2+2)^3+C}



    Edit : a bit slow today... Anti derivative-0010.gif
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  5. #5
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    Question

    This is probably a silly question, but will your answer give the same as mine which is 1/6x^6+x^4+2x^2 + C

    The reason I ask this question is because I believe (i am not 100% sure) that your answer would give a number. what i mean by this is that when you expand your cube, wont a factor of 2 be in the answer.
    I know i didnt do the best job of explaining what I mean.

    Also, could you tell me the formula for expanding a cube.

    Thanks
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  6. #6
    Moo
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    Quote Originally Posted by PensFan10 View Post
    This is probably a silly question, but will your answer give the same as mine which is 1/6x^6+x^4+2x^2 + C

    The reason I ask this question is because I believe (i am not 100% sure) that your answer would give a number. what i mean by this is that when you expand your cube, wont a factor of 2 be in the answer.
    I know i didnt do the best job of explaining what I mean.

    Also, could you tell me the formula for expanding a cube.

    Thanks
    (a+b)^3=a^3+3ab^2+3a^2b+b^3

    It won't give a number
    You will just be able to remove 2^3 in the expansion because C is a constant and so is 2^3
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  7. #7
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    Okay

    The good thing is, it makes sense Thanks
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