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Math Help - Help with indefinite integral substitution

  1. #1
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    Help with indefinite integral substitution

    I'm having trouble figuring out the substitution for \int \frac {5-2\sqrt{x}}{x^3}\frac d{dx}.

    I tried u=x^3 and u=5-2\sqrt{x}, but those don't seem to lead me anywhere.
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  2. #2
    MHF Contributor
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    I assume you mean this:

    \int \frac{5 - 2\sqrt{x}}{x^3} \, dx

    Use the old single term in the denominator trick.

    \int \frac{5}{x^3} - \frac{2}{x^{5/2}} \, dx is a lot easier to work with.
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  3. #3
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    What about \int \frac{x}{x^4+1} \, dx?
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  4. #4
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    Quote Originally Posted by diablo2121 View Post
    What about \int \frac{x}{x^4+1} \, dx?
    This fits a pattern. Look at it as:

    \int\frac{du}{(a^2 + u^2)}

    That fits the pattern for arctan, so:

    u = x^2

    du = 2xdx

    a = 1

    You have to accomodate for the 2 in 2xdx, so:

    \frac{1}{2}\int\frac{2xdx}{1+x^4}

    therefore, the basic integral is:

    \frac{1}{a}\arctan\frac{u}{a}+C


    So from there on it is plugging in numbers! You have to learn these patterns so read up!
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