Results 1 to 5 of 5

Math Help - confused with u-subs problem

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    52

    confused with u-subs problem

    integral of 1/(9 + 4x^2)

    The book says u should be 2/3x. Can someone enlighten me on why this is? I am dumbfounded
    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
    5
    Quote Originally Posted by TYTY View Post
    integral of 1/(9 + 4x^2)

    The book says u should be 2/3x. Can someone enlighten me on why this is? I am dumbfounded
    Have you tried it ....?

    Note that \frac{1}{9 + 4x^2} = \frac{1}{9} \cdot \frac{1}{1 + \left(\frac{2x}{3}\right)^2}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    52
    Quote Originally Posted by mr fantastic View Post
    Have you tried it ....?

    Note that \frac{1}{9 + 4x^2} = \frac{1}{9} \cdot \frac{1}{1 + \left(\frac{2x}{3}\right)^2}.
    please be patient with me here as I still don't understand this. (2x/3)^2 x 9 would have to equal 4x^2 again here right? but wouldn't it be (18x/3)^2 = 6x^2?

    edit: I guess because of the square, you simplify using square root or something. The calculator agrees with your result but somehow it still goes over my head
    Last edited by TYTY; February 22nd 2009 at 06:13 AM.
    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
    5
    Quote Originally Posted by TYTY View Post
    please be patient with me here as I still don't understand this. (2x/3)^2 x 9 would have to equal 4x^2 again here right? but wouldn't it be (18x/3)^2 = 6x^2?
    9 \cdot \left( \frac{2x}{3} \right)^2 = 9 \cdot \frac{2x}{3} \cdot \frac{2x}{3} = 9 \cdot \frac{4x^2}{9} = 4x^2.

    If you're studying calculus it's expected you can do this. And do it easily.

    I suggest you extensively revise basic algebra, index laws etc. if you hope to successfully tackle the sort of calculus questions you're going to get given n the coming days and weeks.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    Posts
    52
    Quote Originally Posted by mr fantastic View Post
    9 \cdot \left( \frac{2x}{3} \right)^2 = 9 \cdot \frac{2x}{3} \cdot \frac{2x}{3} = 9 \cdot \frac{4x^2}{9} = 4x^2.

    If you're studying calculus it's expected you can do this. And do it easily.

    I suggest you extensively revise basic algebra, index laws etc. if you hope to successfully tackle the sort of calculus questions you're going to get given n the coming days and weeks.
    I imagine you're right about needing to review. I haven't taken any classes in 5 years now I find myself in this nightmare situation. Thank you for your explanation though - it did clear it up for me and now in retrospect it does seem kind of obvious.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. u subs and x bounds on integrals
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 5th 2009, 01:44 PM
  2. [SOLVED] trig subs
    Posted in the Calculus Forum
    Replies: 12
    Last Post: April 5th 2009, 12:43 PM
  3. Is this u-subs?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 11th 2009, 10:32 AM
  4. Subs
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 30th 2007, 07:43 PM
  5. Help with trig subs...
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 5th 2007, 04:33 PM

Search Tags


/mathhelpforum @mathhelpforum