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Math Help - Can somebody help me understand why for this indefinite integral?

  1. #1
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    Can somebody help me understand why for this indefinite integral?

    I am being asked to find the indefinite integral
    \int x(x^{2}-1)^3dx
    The solution reads as:
    The factor x in the integrand is, except for a constant multiple, the derivative of x^{2}-1,
    so that we can apply equation
    \int[f(x)]^{n}f'(x)dx=\frac{1}{n+1}[f(x)]^{n+1}+c
    with f(x)=x^{2}-1 and n=3.
    Since f'(x)=2x, we write x=\frac{1}{2}(2x) before applying the formula.
    Thus we have
    \int x(x^{2}-1)^{3}dx=\frac{1}{2}\int(x^{2}-1)^{3}(2x)dx
    Why the (2x) and why the \frac{1}{2} outside of the integral?
    Many thanks all.
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  2. #2
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    Re: Can somebody help me understand why for this indefinite integral?

    Because (x^2-1)^3(2x) precisely equals [f(x)]^3f'(x) for f(x)=x^2-1. We want to represent the integrand in this form because, as the text says,

    \int[f(x)]^{n}f'(x)dx=\frac{1}{n+1}[f(x)]^{n+1}+c
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  3. #3
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    Re: Can somebody help me understand why for this indefinite integral?

    So if I see a question like this, I should find the derivative of what's inside the brackets, if it matches what's outside the brackets (but has a numerical multiple) then I use the equation above, but include a fraction to counter the numerical multiple so it equals 1? And the numerical fraction (in this case a half to cancel the 2 to a 1) goes outside the integral to comply with the constant multiple rule?

    (struggling a little here)
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  4. #4
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    Re: Can somebody help me understand why for this indefinite integral?

    The penny has dropped, I get it now,
    many thanks.
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  5. #5
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    Re: Can somebody help me understand why for this indefinite integral?

    Yes. In general, you massage the expression in whatever way to give it a form for which one knows how to find an integral.
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  6. #6
    Senior Member bugatti79's Avatar
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    Re: Can somebody help me understand why for this indefinite integral?

    Quote Originally Posted by emakarov View Post
    Because (x^2-1)^3(2x) precisely equals [f(x)]^3f'(x) for f(x)=x^2-1. We want to represent the integrand in this form because, as the text says,

    \int[f(x)]^{n}f'(x)dx=\frac{1}{n+1}[f(x)]^{n+1}+c
    Would I be right in saying this is the substitution method i.e. the above LHS is equivalent to

    \int f(g(x))g'(x)dx where f(g(x)) would be equal to [f(x)]^{n}

    I get the same answer with this version. Thanks
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  7. #7
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    Re: Can somebody help me understand why for this indefinite integral?

    Yes, this is the substitution method.
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    Senior Member bugatti79's Avatar
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    Re: Can somebody help me understand why for this indefinite integral?

    Quote Originally Posted by emakarov View Post
    Yes, this is the substitution method.
    Would I be correct in saying that is the product form of the method. Would the quotient form be the inverse i.e.

    \int [f(x)]^{-n} f'(x) dx or \int [f(x)]^{n} f'(x)^{-1} dx

    Would these work? Thanks
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  9. #9
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    Re: Can somebody help me understand why for this indefinite integral?

    Quote Originally Posted by bugatti79 View Post
    Would I be correct in saying that is the product form of the method. Would the quotient form be the inverse i.e.

    \int [f(x)]^{-n} f'(x) dx or \int [f(x)]^{n} f'(x)^{-1} dx

    Would these work? Thanks
    no ... in both cases of substitution, if u = f(x) , then the "du" is f'(x) dx
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  10. #10
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    Re: Can somebody help me understand why for this indefinite integral?

    Quote Originally Posted by bugatti79 View Post
    Would the quotient form be the inverse i.e.

    \int [f(x)]^{-n} f'(x) dx or \int [f(x)]^{n} f'(x)^{-1} dx
    Suppose that G'(y)=g(y) for some functions G(y), g(y). Then \int g(f(x))f'(x)\,dx=\int [G(f(x))]'\,dx=G(f(x))+C. This works whether g(y) is y^n or y^{-n}.

    So, \int [f(x)]^{-n} f'(x) dx=\frac{f(x)^{-n+1}}{-n+1}+C for n\ne 1. Here g(y)=y^{-n} and G(y)=\frac{y^{-n+1}}{-n+1}.

    I don't see how to apply the substitution method to \int [f(x)]^{n} f'(x)^{-1} dx.
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