Results 1 to 4 of 4

Math Help - help! integrals!

  1. #1
    Junior Member
    Joined
    Jan 2008
    Posts
    26

    help! integrals! - URGENT

    integral of:
    3(x^4/e^x^5)dx

    I'm pretty sure its substitution rule but I don't know what to put as the u

    u=x^4?
    u=e^x^5 help!

    and:

    integral of:
    5e^s/1+3e^s
    Last edited by s0urgrapes; March 6th 2008 at 01:29 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Dec 2007
    Posts
    131
    {\text{Let }}u = e^{x^5 }\text{  ;  }<br />
  du = 5x^4 e^{x^5 }\text{ }  dx


    \int {3\frac{{x^4 }}<br />
{{e^{x^5 } }}} {\text{ }}dx = \frac{3}<br />
{5}\int {\frac{{du}}<br />
{{u^2 }}}  = \frac{3}<br />
{5}\left( { - \frac{1}<br />
{u}} \right) =  - \frac{3}<br />
{{5u}} =  - \frac{3}<br />
{{5e^{x^5 } }}<br />
 = \boxed{ - \frac{{3e^{ - x^5 } }}<br />
{5}}<br />


    {\text{Let }}u =1 + 3e^{s}\text{  ;  }<br />
  du = 3e^{s}\text{ } ds


    \int {\frac{{5e^s }}<br />
{{1 + 3e^s }}} {\text{ }}ds = \frac{5}<br />
{3}\int {\frac{{du}}<br />
{u}}  = \frac{5}<br />
{3}\ln \left( {\left| u \right|} \right) = \frac{5}<br />
{3}\ln \left( {\left| {1 + 3e^s } \right|} \right) = \frac{5}<br />
{3}\ln \left( {1 + 3e^s } \right)<br /> <br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2008
    Posts
    26
    Quote Originally Posted by xifentoozlerix View Post
    {\text{Let }}u = e^{x^5 }\text{  ;  }<br />
  du = 5x^4 e^{x^5 }\text{ }  dx


    \int {3\frac{{x^4 }}<br />
{{e^{x^5 } }}} {\text{ }}dx = \frac{3}<br />
{5}\int {\frac{{du}}<br />
{{u^2 }}}  = \frac{3}<br />
{5}\left( { - \frac{1}<br />
{u}} \right) =  - \frac{3}<br />
{{5u}} =  - \frac{3}<br />
{{5e^{x^5 } }}<br />
 = \boxed{ - \frac{{3e^{ - x^5 } }}<br />
{5}}<br />


    {\text{Let }}u =1 + 3e^{s}\text{  ;  }<br />
  du = 3e^{s}\text{ } ds


    \int {\frac{{5e^s }}<br />
{{1 + 3e^s }}} {\text{ }}ds = \frac{5}<br />
{3}\int {\frac{{du}}<br />
{u}}  = \frac{5}<br />
{3}\ln \left( {\left| u \right|} \right) = \frac{5}<br />
{3}\ln \left( {\left| {1 + 3e^s } \right|} \right) = \frac{5}<br />
{3}\ln \left( {1 + 3e^s } \right)<br /> <br />
    thanks! just a question, how do you get du/u^2 in the first one, how come it isn't du/u
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2007
    Posts
    131
    Well, let's see what happens if you do \int {\frac{{du}}{u}}:


    \int {\frac{{du}}<br />
{u}}  = \int {\frac{{5x^4 e^{x^5 } {\text{ }}dx}}<br />
{{e^{x^5 } }}}  = \int {5x^4 dx} <br />

    As you can see, the term {e^{x^5 } } cancels and you end up with a different integral. With {u^2 } however, the top {e^{x^5 } } in the numerator is taken care of, and it leaves the {e^{x^5 } } in the denominator as the original problem had:


    \int {\frac{{du}}<br />
{{u^2 }}}  = \int {\frac{{5x^4 e^{x^5 } {\text{ }}dx}}<br />
{{\left( {e^{x^5 } } \right)^2 }}}  = \int {\frac{{5x^4 dx}}<br />
{{e^{x^5 } }}}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Contour Integrals (to Evaluate Real Integrals)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: January 17th 2011, 10:23 PM
  2. Replies: 1
    Last Post: December 6th 2009, 08:43 PM
  3. Integrals : 2
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 24th 2009, 08:40 AM
  4. Integrals and Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 9th 2009, 05:52 PM
  5. integrals Help please
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 8th 2008, 07:16 PM

Search Tags


/mathhelpforum @mathhelpforum