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Thread: 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
    $\displaystyle \int x(x^{2}-1)^3dx$
    The solution reads as:
    The factor $\displaystyle x$ in the integrand is, except for a constant multiple, the derivative of $\displaystyle x^{2}-1$,
    so that we can apply equation
    $\displaystyle \int[f(x)]^{n}f'(x)dx=\frac{1}{n+1}[f(x)]^{n+1}+c$
    with $\displaystyle f(x)=x^{2}-1$ and $\displaystyle n=3$.
    Since $\displaystyle f'(x)=2x$, we write $\displaystyle x=\frac{1}{2}(2x)$ before applying the formula.
    Thus we have
    $\displaystyle \int x(x^{2}-1)^{3}dx=\frac{1}{2}\int(x^{2}-1)^{3}(2x)dx$
    Why the $\displaystyle (2x)$ and why the $\displaystyle \frac{1}{2}$ 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 $\displaystyle (x^2-1)^3(2x)$ precisely equals $\displaystyle [f(x)]^3f'(x)$ for $\displaystyle f(x)=x^2-1$. We want to represent the integrand in this form because, as the text says,

    $\displaystyle \int[f(x)]^{n}f'(x)dx=\frac{1}{n+1}[f(x)]^{n+1}+c$
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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 $\displaystyle (x^2-1)^3(2x)$ precisely equals $\displaystyle [f(x)]^3f'(x)$ for $\displaystyle f(x)=x^2-1$. We want to represent the integrand in this form because, as the text says,

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

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

    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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  8. #8
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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.

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

    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.

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

    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.

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

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

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