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    Substitution u = x^2-4
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    (Original post by DFranklin)
    Substitution u = x^2-4
    Thanks so much but can you not do it this way:Name:  7.PNG
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    I understand the substitution way by the way.

    EDIT : Sorry I asked a silly question here You can't always use a certain integration technique. Thanks so much for the help
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    (Original post by sienna2266)
    Thanks so much but can you not do it this way:Name:  7.PNG
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    No - the solution doesn't involve ln, so your formula can't possibly apply.

    I understand the substitution way by the way.
    Note that your f'(x)/f(x) example is merely the same thing as applying the substitution u = f(x); the f'(x) is divided out and you are left with  \int \frac{1}{u}\, du, which comes to ln|u| = ln|f(x)|.

    Try integrating \frac{x}{x^2-4}\,dx using the substitution u = x^2-4; you should be able to see how this isn't really any different from the formula you're using, it's just there are some steps your formula 'skips' out.

    The point is that you can use the same approach (if not formula) for *any* scenario where you have \int f'(x) g(f(x))\, dx and you know how to integrate \int g(u) \,du.

    More explictly, if \int g(u)\,du = G(u) + C, then \int f'(x) g(f(x)) \, dx = G(f(u)) + C

    In this case, you have g(u) = u^{-1/2}, rather than g(u) = 1/u in the formula you've quoted. But g is still easy to integrate, so you can do this all mentally with practice.
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    (Original post by DFranklin)
    No - the solution doesn't involve ln, so your formula can't possibly apply.

    Note that your f'(x)/f(x) example is merely the same thing as applying the substitution u = f(x); the f'(x) is divided out and you are left with  \int \frac{1}{u}\, du, which comes to ln|u| = ln|f(x)|.

    Try integrating \frac{x}{x^2-4}\,dx using the substitution u = x^2-4; you should be able to see how this isn't really any different from the formula you're using, it's just there are some steps your formula 'skips' out.

    The point is that you can use the same approach (if not formula) for *any* scenario where you have \int f'(x) g(f(x))\, dx and you know how to integrate \int g(u) \,du.

    More explictly, if \int g(u)\,du = G(u) + C, then \int f'(x) g(f(x)) \, dx = G(f(u)) + C

    In this case, you have g(u) = u^{-1/2}, rather than g(u) = 1/u in the formula you've quoted. But g is still easy to integrate, so you can do this all mentally with practice.
    Thank you so much!
 
 
 
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