Results 1 to 3 of 3

Thread: Integration

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    36

    Cool Integration

    Integrate (3+x)/(x^2 -9)^0.5

    For now, i would just like to know if i am supposed to use the by parts forumla?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701
    Quote Originally Posted by Erghhh View Post
    Integrate (3+x)/(x^2 -9)^0.5

    For now, i would just like to know if i am supposed to use the by parts forumla?
    $\displaystyle \int \frac{x}{\sqrt{x^2-9}} \, dx + \int \frac{3}{\sqrt{x^2-9}} \, dx$

    straight up substitution for the first integral , $\displaystyle u = x^2 - 9$

    trig sub for the second, $\displaystyle x = 3\sec{t}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848

    Post

    Hello, Erghhh!

    I'd use a trig substitution . . .


    $\displaystyle \int \frac{x+3}{\sqrt{x^2-9}}\,dx$
    Let: .$\displaystyle x \:=\:3\sec\theta \quad\Rightarrow\quad dx \:=\:3\sec\theta\tan\theta\,d\theta$
    and: .$\displaystyle \sqrt{x^2-9} \:=\:3\tan\theta$

    Substitute: .$\displaystyle \int \frac{3\sec\theta + 3}{3\tan\theta}(3\sec\theta\tan\theta\,d\theta) \;=\;3\int(\sec\theta + 1)(\sec\theta\,d\theta)$

    . . $\displaystyle =\;3\int(\sec^2\!\theta + \sec\theta)\,d\theta \;=\;3\bigg[\tan\theta + \ln|\sec\theta + \tan\theta| \bigg] + C $


    Back-substitute: .$\displaystyle \sec\theta \:=\:\frac{x}{3} \:=\:\frac{hyp}{adj}$
    . . Then: .$\displaystyle opp = \sqrt{x^2-9} \quad\Rightarrow\quad \tan\theta \:=\:\frac{\sqrt{x^2-9}}{3}$


    We have: .$\displaystyle 3\left[\frac{\sqrt{x^2-9}}{3} + \ln\left|\frac{x}{3} + \frac{\sqrt{x^2-9}}{3}\right|\,\right] + C$

    . . . . . . . $\displaystyle = \;\sqrt{x^2-9} + 3\ln\left|\frac{x+\sqrt{x^2-9}}{3}\right| + C $

    . . . . . . . $\displaystyle = \;\sqrt{x^2-9} + 3\ln\left|x + \sqrt{x^2-9}\right| + C $ .**


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    ** .What happened to the denominator, you ask?
    . . . . Watch this!


    $\displaystyle \sqrt{x^2-9} \:+\: 3\ln\left|\frac{x + \sqrt{x^2-9}}{3}\right| \:+\: C$

    . . $\displaystyle = \;\;\sqrt{x^2-9} \:+\: 3\bigg[\ln\left|x + \sqrt{x^2-9}\right| - \ln 3\bigg] \:+\: C $

    . . $\displaystyle = \;\;\sqrt{x^2-9} \:+\: 3\ln\left|x + \sqrt{x^2-9}\right| \:-\: \underbrace{3\ln 3 \:+\: C}_{\text{a constant!}}$

    . . $\displaystyle = \;\;\sqrt{x^2-9} \:+\:3\ln\left|x + \sqrt{x^2-9}\right| + C$


    Textbooks love to pull this on us . . . You've been warned!

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Nov 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: Nov 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: Sep 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: Feb 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum