Results 1 to 7 of 7

Thread: Simple Integration

  1. #1
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1

    Simple Integration

    $\displaystyle \int \frac{u}{\sqrt{u^2 +16}} \ du$

    Thanks in advance.
    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 Air View Post
    $\displaystyle \int \frac{u}{\sqrt{u^2 +16}} \ du$

    Thanks in advance.
    Make the substitution $\displaystyle w = u^2 + 16$. Note that $\displaystyle \frac{dw}{du} = 2u \Rightarrow du = \frac{dw}{2u}$ ....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi

    $\displaystyle \frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)=?$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by flyingsquirrel View Post
    Hi

    $\displaystyle \frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)=?$
    $\displaystyle \frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right) = \frac{ \mathrm{d}}{\mathrm{d}u} \left( u^2+16 \right)^\frac{1}{2} = \frac{1}{2}(u^2 + 16)^{-\frac{1}{2}}(2u) = (u)(u^2 + 16)^{-\frac{1}{2}} = \frac{u}{\sqrt{u^2+16}}$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by mr fantastic View Post
    Make the substitution $\displaystyle w = u^2 + 16$. Note that $\displaystyle \frac{dw}{du} = 2u \Rightarrow du = \frac{dw}{2u}$ ....
    You should try this way to see why flyingsquirrel was able to suggest you consider the derivative $\displaystyle \frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)$ ......
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by mr fantastic View Post
    You should try this way to see why flyingsquirrel was able to suggest you consider the derivative $\displaystyle \frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)$ ......
    $\displaystyle \int \frac{u}{\sqrt{u^2 + 16}} \ du$

    $\displaystyle w = u^2 + 16 \therefore \frac{dw}{du} = 2u \implies du = \frac{dw}{2u}$

    $\displaystyle \int \frac{u}{\sqrt w} . \frac{dw}{2u}$

    $\displaystyle \int \frac{1}{\sqrt w} . \frac{dw}{2}$

    $\displaystyle \frac{1}{2} \int \frac{1}{\sqrt w} \ dw$

    $\displaystyle \frac{1}{2} \left[ 2 w ^{\frac{1}{2}} \right] $

    $\displaystyle w^{\frac{1}{2}} = (u^2+16)^\frac{1}{2} = \sqrt{u^2 + 16}$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by Air View Post
    $\displaystyle \int \frac{u}{\sqrt{u^2 + 16}} \ du$

    $\displaystyle w = u^2 + 16 \therefore \frac{dw}{du} = 2u \implies du = \frac{dw}{2u}$

    $\displaystyle \int \frac{u}{\sqrt w} . \frac{dw}{2u}$

    $\displaystyle \int \frac{1}{\sqrt w} . \frac{dw}{2}$

    $\displaystyle \frac{1}{2} \int \frac{1}{\sqrt w} \ dw$

    $\displaystyle \frac{1}{2} \left[ 2 w ^{\frac{1}{2}} \right] $

    $\displaystyle w^{\frac{1}{2}} = (u^2+16)^\frac{1}{2} = \sqrt{u^2 + 16}$
    + C.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simple integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 20th 2011, 12:36 PM
  2. simple integration
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Jun 14th 2011, 04:07 AM
  3. Replies: 2
    Last Post: Feb 19th 2010, 10:55 AM
  4. Simple Integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jan 31st 2009, 04:32 PM
  5. simple integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 20th 2008, 04:49 PM

Search Tags


/mathhelpforum @mathhelpforum