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Math Help - integration by substitution...quotient

  1. #1
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    integration by substitution...quotient

    i have no idea whats going on..please someone explain..it would be great..i get the starting steps but i get lost in the middle...

    integrate 5x^3 / 1 + 16x^4

    what i have so far is...

    Let u = 1 + 16x^4
    du/dx = 64x^3
    du = 64x^3 dx

    now what do i do??

    i know that it has to be 5/u and then something.....
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  2. #2
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    Quote Originally Posted by b00yeah05 View Post
    i have no idea whats going on..please someone explain..it would be great..i get the starting steps but i get lost in the middle...

    integrate 5x^3 / 1 + 16x^4

    what i have so far is...

    Let u = 1 + 16x^4
    du/dx = 64x^3
    du = 64x^3 dx

    now what do i do??

    i know that it has to be 5/u and then something.....
    The good news is that you're almost on exactly the right track.

    du = 64x^3 dx \Rightarrow dx = \frac{du}{64x^3}

    So the integral becomes \int \frac{5x^3}{u} \, \frac{du}{64x^3} = \int \frac{5}{64} \, \frac{1}{u} \, du = \frac{5}{64} \int \frac{1}{u} \, du = ......
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    yes..thanks mr fantastic i worked that one out eventually..but just a general question on integration by substitution now.. do you always have to times by 'du' in the end? for all questions??
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    Quote Originally Posted by b00yeah05 View Post
    yes..thanks mr fantastic i worked that one out eventually..but just a general question on integration by substitution now.. do you always have to times by 'du' in the end? for all questions??
    You have to substitute for x AND dx. The expression for dx will always contain a du. So yes, I guess that means you will always have to "times by a du in the end" ...
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  5. #5
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    Quote Originally Posted by b00yeah05 View Post
    i have no idea whats going on..please someone explain..it would be great..i get the starting steps but i get lost in the middle...

    integrate 5x^3 / 1 + 16x^4

    what i have so far is...

    Let u = 1 + 16x^4
    du/dx = 64x^3
    du = 64x^3 dx

    now what do i do??

    i know that it has to be 5/u and then something.....
    You should use parentheses, your question looks like \int \frac{5x^3}1 + 16x^4 So maybe next time write it as 5x^3 /(1 + 16x^4)

    Anyway,
    \int \frac{5x^3}{1 + 16x^4}dx

    5 is a coefficient, so you can pull it to the outside
    =5\int \frac{x^3}{1 + 16x^4}dx

    Now partition your integral so that it is easier to see what you are doing
    \begin{array}{rcc}<br />
=5\int &\frac{1}{1 + 16x^4}&x^3~dx\\<br />
\end{array}

    \begin{array}{|c|}<br />
\hline SUBSTITUTION\\\hline\\<br />
 u=1+16x^4\\\\<br />
du=64x^3dx\\\\<br />
\frac 1{64}du=x^3dx\\\\\hline\end{array}


    Determine what will substitute with what
    \begin{array}{rcc}<br />
&&\frac 1{64}~du\\<br />
=5\int &\underbrace{\frac{1}{1 + 16x^4}}&\overbrace{x^3~dx}\\<br />
&u<br />
\end{array}

    So make the substitution
    =5\int \frac{1}{u}*\frac{1}{64}~du

    1/64 is a constant, so it is part of the coefficient, so move it to the front to get it out of your way.
    =5*\frac{1}{64}\int \frac{1}{u}~du

    =\frac{5}{64}\int \frac{1}{u}~du

    Now, you should have memorized that \int \frac 1x ~dx = ln|x|+C so apply this:

    =\frac{5}{64}~ln|u|+C

    And now reverse substitute from u back into x:
    =\frac{5}{64}~ln|1+16x^4|+C


    edit: beat me to it >.< Darn arrays
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  6. #6
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    Quote Originally Posted by b00yeah05 View Post
    i have no idea whats going on..please someone explain..it would be great..i get the starting steps but i get lost in the middle...

    integrate 5x^3 / 1 + 16x^4
    Based on the idea where \int {\frac{{f'(x)}}<br />
{{f(x)}}\,dx}  = \ln \left| {f(x)} \right| + k,\,f(x) \ne 0, you just turn the integral into that form:

    \int {\frac{{5x^3 }}<br />
{{1 + 16x^4 }}\,dx}  = \frac{5}<br />
{{64}}\int {\frac{{\left( {1 + 16x^4 } \right)'}}<br />
{{1 + 16x^4 }}\,dx}  = \frac{5}<br />
{{64}}\ln \left( {1 + 16x^4 } \right) + k.

    (Absolute value is omitted here since 1+16x^4 it's always positive.)

    P.S.: see my signature for better LaTeX typesetting.
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