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Math Help - Integration by substitution

  1. #1
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    Integration by substitution

    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
    Integrate: x root (4x-3) using substitution u = 4x-3
    do as they suggest

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

    \Rightarrow du = 4~dx

    \Rightarrow \frac 14 ~du = dx

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

    \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

    \int (u + 3) \sqrt u~du
    Am I able to expand (u + 3) \sqrt u~du os that it becomes 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 (u + 3) \sqrt u~du os that it becomes u\sqrt u + 3\sqrt u? Does it work like that?
    Yes. And then write u\sqrt u + 3\sqrt u = u^{3/2} + 3 u^{1/2}
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