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Thread: Integration by substitution

  1. #1
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    Integration by substitution

    $\displaystyle Integrate: x root (4x-3)$ using substitution u = 4x-3

    Sorry, don't know how to use the codes to make it look mathematical. Where can I learn to do that?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nerdzor View Post
    $\displaystyle Integrate: x root (4x-3)$ using substitution u = 4x-3
    do as they suggest

    Let $\displaystyle u = 4x - 3 \implies \boxed{x = \frac 14(u + 3)}$

    $\displaystyle \Rightarrow du = 4~dx$

    $\displaystyle \Rightarrow \frac 14 ~du = dx$

    By plugging in all these pieces, we obtain the integral,

    $\displaystyle \frac 1{16} \int (u + 3) \sqrt u~du$

    now continue, hopefully you see how we got to that

    Sorry, don't know how to use the codes to make it look mathematical. Where can I learn to do that?
    here
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  3. #3
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    Quote Originally Posted by Jhevon View Post

    $\displaystyle \int (u + 3) \sqrt u~du$
    Am I able to expand $\displaystyle (u + 3) \sqrt u~du$ os that it becomes $\displaystyle u\sqrt u + 3\sqrt u$? Does it work like that?
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  4. #4
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    Quote Originally Posted by nerdzor View Post
    Am I able to expand $\displaystyle (u + 3) \sqrt u~du$ os that it becomes $\displaystyle u\sqrt u + 3\sqrt u$? Does it work like that?
    Yes. And then write $\displaystyle u\sqrt u + 3\sqrt u = u^{3/2} + 3 u^{1/2}$
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