Results 1 to 8 of 8

Math Help - [SOLVED] Calculus: Integration by substitution

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    74

    Question [SOLVED] Calculus: Integration by substitution

    I'm stuck...Could someone please help explain how to do this problem to me?

    Find by substitution: the integral e to e^2 of dx/x(lnx)^2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    \int_e^{e^2 } {\frac{{dx}}<br />
{{x\left( {\ln (x)} \right)^2 }}}  = \left. {\frac{{ - 1}}<br />
{{\ln (x)}}} \right|_e^{e^2 }
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by juicysharpie View Post
    I'm stuck...Could someone please help explain how to do this problem to me?

    Find by substitution: the integral e to e^2 of dx/x(lnx)^2
    Just substitute t=\ln(x)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2009
    Posts
    74
    I'm sorry. I still don't understand...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    It's basic stuff according to substitution method. What is exactly you don't get?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2009
    Posts
    74

    Unhappy

    I found u=ln x. du/dx = 1/x and du= 1/x dx. I found the new limits, 1 (lower) 2 (upper). So the new equation is the integral from 1 to 2 of 1/u^2 du. How do I derivate that so that I can plug in 1 and 2 to solve? I tried ln u x ln u but I don't think that's correct because I did not get 1/2 like the answer is supposed to be...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by juicysharpie View Post
    I found u=ln x. du/dx = 1/x and du= 1/x dx. I found the new limits, 1 (lower) 2 (upper). So the new equation is the integral from 1 to 2 of 1/u^2 du.
    \int_1^2 {\frac{{du}}<br />
{{u^2 }}}  = \int_1^2 {u^{ - 2} du} <br />
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jan 2009
    Posts
    74

    Thumbs up

    oh, i see...thanks for the help!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 22nd 2010, 08:03 AM
  2. [SOLVED] Integration by substitution
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 2nd 2010, 10:08 PM
  3. [SOLVED] Integration by Substitution 2
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 14th 2009, 08:22 PM
  4. [SOLVED] Integration by Substitution
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 14th 2009, 06:27 PM
  5. [SOLVED] Integration by Substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 11th 2009, 01:28 PM

Search Tags


/mathhelpforum @mathhelpforum