Results 1 to 6 of 6

Thread: integration by substitution...quotient

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    91

    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 $\displaystyle 5x^3 / 1 + 16x^4$

    what i have so far is...

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

    now what do i do??

    i know that it has to be 5/u and then something.....
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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 $\displaystyle 5x^3 / 1 + 16x^4$

    what i have so far is...

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

    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.

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

    So the integral becomes $\displaystyle \int \frac{5x^3}{u} \, \frac{du}{64x^3} = \int \frac{5}{64} \, \frac{1}{u} \, du = \frac{5}{64} \int \frac{1}{u} \, du = ......$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2008
    Posts
    91
    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??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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" ...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    724
    Awards
    1
    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 $\displaystyle 5x^3 / 1 + 16x^4$

    what i have so far is...

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

    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 $\displaystyle \int \frac{5x^3}1 + 16x^4$ So maybe next time write it as $\displaystyle 5x^3 /(1 + 16x^4)$

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

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

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

    $\displaystyle \begin{array}{|c|}
    \hline SUBSTITUTION\\\hline\\
    u=1+16x^4\\\\
    du=64x^3dx\\\\
    \frac 1{64}du=x^3dx\\\\\hline\end{array}$


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

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

    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.
    $\displaystyle =5*\frac{1}{64}\int \frac{1}{u}~du$

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

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

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

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


    edit: beat me to it >.< Darn arrays
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    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 $\displaystyle 5x^3 / 1 + 16x^4$
    Based on the idea where $\displaystyle \int {\frac{{f'(x)}}
    {{f(x)}}\,dx} = \ln \left| {f(x)} \right| + k,\,f(x) \ne 0,$ you just turn the integral into that form:

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

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

    P.S.: see my signature for better LaTeX typesetting.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. quotient integration!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 27th 2009, 01:23 PM
  2. integration with substitution HELP!
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Sep 25th 2009, 05:23 AM
  3. Integration by substitution
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Sep 19th 2009, 08:54 AM
  4. Integration by u substitution
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 14th 2008, 02:33 PM
  5. integration of quotient
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Sep 16th 2008, 04:16 AM

Search Tags


/mathhelpforum @mathhelpforum