Results 1 to 7 of 7

Math Help - Evaluating Integrals using the Natural Log function? really lost

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    23

    Evaluating Integrals using the Natural Log function? really lost

    Bare with me, I just registered so I have no idea how to type any signs ... difficult ones will be written out :o)sorry and thanks.

    The problem: The integral from 1-e ((x^2+x+1)/x) dx

    So far ... I got u= x^2+x+1 ; du= 2x+1 dx ; -1/2du= xdx <-- i'm not quite sure about this

    Integral (1-e) ((x^2+x+1)/x) dx = -1/2 Integral (0-1) (u/du) ... And this is where I get stuck. Help please. Thnks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2009
    Posts
    112
    Is this your integral?

    \int 1-e^{\frac{x^2+x+1}{x}}dx
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
     \int_1^e\frac{x^2+x+1}{x}dx =\int_1^e\frac{x^2}{x}+\frac{x}{x}+\frac{1}{x}dx = \int_1^ex+1+\frac{1}{x}dx = \int_1^ex~dx+\int_1^e1~dx+\int_1^e\frac{1}{x}~dx
    Last edited by pickslides; September 9th 2009 at 04:01 PM. Reason: bad latex
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2009
    Posts
    23
    Quote Originally Posted by seld View Post
    Is this your integral?

    \int 1-e^{\frac{x^2+x+1}{x}}dx
    yes seld it is
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2009
    Posts
    112
    \int 1-e^{\frac{x^2+x+1}{x}}dx

    You can certainly distribute the x as pick showed you:
    so your integral then becomes:

    \int 1-e^{x+1+\frac{1}{x}}dx


    can you take it from there?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2009
    Posts
    23
    Quote Originally Posted by seld View Post
    \int 1-e^{\frac{x^2+x+1}{x}}dx

    You can certainly distribute the x as pick showed you:
    so your integral then becomes:

    \int 1-e^{x+1+\frac{1}{x}}dx


    can you take it from there?
    Yeah I started doing it that way which is a lot easier but I got
    {x^2/2+x+ln abs(x)} - now I just have to plug in e and 1 into the variable right? My integral looks the way seld wrote it out, I think I miss informed you when I said my integral was the way you asked. sorry.
    Last edited by bgonzal8; September 9th 2009 at 04:23 PM. Reason: messed up
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2009
    Posts
    112
    This might be a little above your level, maybe not.

    So you have:

    \int 1-e^{x+1+\frac{1}{x}}dx

    You can split off the 1.

    \int 1 dx + \int -e*e^{x + \frac{1}{x}}dx

    so you pull the constant -e out.
    \int 1 dx + -e\int e^{x + \frac{1}{x}}dx

    This is where it starts to gets tricky. So the way I solved it was I used the fact that this is true: e^x=\sum_{k=0}^\infty\frac{x^k}{k!}


    So plugging in the x + 1/x you get:

    e^{x+\frac{1}{x}}=\sum_{k=0}^\infty\frac{(x+\frac{  1}{x})^k}{k!}

    So then you have:

    \int 1 dx + -e\int\sum_{k=0}^\infty\frac{(x+\frac{1}{x})^k}{k!}  dx

    Since the integral of a sum is the sum of the integral you can move that outside of the integral.


    \int 1 dx + -e\sum_{k=0}^\infty\int\frac{(x+\frac{1}{x})^k}{k!}  dx

    \int 1 dx + -e\sum_{k=0}^\infty\frac{1}{k!}\int(x+\frac{1}{x})^  kdx

    Then (x+1/x)^k is hard to work with but we have binomial theorem which states: (a+b)^n=\sum_{i=0}^n {n \choose i}a^ib^{n-i}

    So you plug x and 1/x into binomial theorem and you have:

    \int 1 dx + -e\sum_{k=0}^\infty\frac{1}{k!}\int\sum_{i=0}^k{k \choose i}x^i\frac{1}{x}^{k-i}dx

    You move the sum out again:

    \int 1 dx + -e\sum_{k=0}^\infty\frac{1}{k!}\sum_{i=0}^k\int {k \choose i}x^i\frac{1}{x}^{k-i}dx

    the {k \choose i} is a constant, it's also known as k choose i so you can pull it out.

    \int 1 dx + -e\sum_{k=0}^\infty\frac{1}{k!}\sum_{i=0}^k {k \choose i}\int \frac{x^i}{x^{k-i}}dx

    So rewriting the term you get:

    \int 1 dx + -e\sum_{k=0}^\infty\frac{1}{k!}\sum_{i=0}^k {k \choose i}\int x^{2i-k}dx

    and lastly power rule:

    x + -e\sum_{k=0}^\infty\frac{1}{k!}\sum_{i=0}^k {k \choose i}\frac{x^{2i-k+1}}{2i-k+1} + C

    but that's the solution, it's not . . . pretty though I'm not sure if there's a prettier way to write it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Evaluating integrals
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 29th 2010, 09:16 PM
  2. Evaluating integrals
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 14th 2010, 05:13 PM
  3. Evaluating the Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 9th 2009, 05:51 PM
  4. Replies: 5
    Last Post: April 27th 2009, 09:16 AM
  5. Some help with evaluating integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 23rd 2008, 04:20 PM

Search Tags


/mathhelpforum @mathhelpforum