Results 1 to 7 of 7

Math Help - How do you solve this integral?

  1. #1
    Junior Member
    Joined
    Feb 2008
    Posts
    25

    How do you solve this integral?

    \int_{1}^{2} 1/(x+1)^2 \: dx

    not sure how to solve an integral with a quantity squared in the denominator. I'm new to integration. do you foil the bottom or rewrite it as (x+1)^-2? or what?

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by mojo0716 View Post
    \int_{1}^{2} 1/(x+1)^2 \: dx
    Make the u substitution u=x+1\implies \,du=\,dx

    Thus, the integral transforms into \int_{1}^{2} \frac{\,dx}{(x+1)^2}\implies\int_2^3\frac{\,du}{u^  2}

    I'm sure you can take it from here.

    --Chris
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2008
    Posts
    25
    Quote Originally Posted by Chris L T521 View Post
    Make the u substitution u=x+1\implies \,du=\,dx

    Thus, the integral transforms into \int_{1}^{2} \frac{\,dx}{(x+1)^2}\implies\int_2^3\frac{\,du}{u^  2}

    I'm sure you can take it from here.

    --Chris
    ah thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2008
    Posts
    25
    The solution manual has it like this, though, and I wonder what they did? Looks like they didn't use any substitution.

    \int_{1}^{2} 1/(x+1)^2 \: dx = (-1/(x+1)) from 1 to 2

    after that I know what to do, but I don't know how they got = (-1/(x+1)) from 1 to 2

    thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by mojo0716 View Post
    The solution manual has it like this, though, and I wonder what they did?

    \int_{1}^{2} 1/(x+1)^2 \: dx = (-1/x+1) from 1 to 2



    thanks
    There way is correct, and so is mine.

    When we integrate \frac{1}{(x+1)^2}, we still need to apply a substitution.

    The integral would become \int\frac{\,du}{u^2}=-\frac{1}{u}

    What they did was re-substitute u back into the antiderivative:

    -\frac{1}{u}=-\frac{1}{x+1}, and then they evaluated from 1 to 2.

    However, the way I did it, I eliminated the need to re-substitute u back into the equation.

    I did this when I converted the integration limits.

    Since u=x+1 and the x limits of integration were from 1 to 2, I saw that the u limits would be from u=1+1=\color{red}\bold 2 to u=2+1=\color{red}\bold 3

    So when you evaluate, we see that

    \int_1^2\frac{\,dx}{(x+1)^2}=\int_2^3\frac{\,du}{u  ^2}=\color{red}\boxed{\frac{1}{6}}

    I hope this clarified things! If you have questions, feel free to ask. We're here to help you!

    --Chris
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2008
    Posts
    25
    Quote Originally Posted by Chris L T521 View Post
    There way is correct, and so is mine.

    When we integrate \frac{1}{(x+1)^2}, we still need to apply a substitution.

    The integral would become \int\frac{\,du}{u^2}=-\frac{1}{u}

    What they did was re-substitute u back into the antiderivative:

    -\frac{1}{u}=-\frac{1}{x+1}, and then they evaluated from 1 to 2.

    However, the way I did it, I eliminated the need to re-substitute u back into the equation.

    I did this when I converted the integration limits.

    Since u=x+1 and the x limits of integration were from 1 to 2, I saw that the u limits would be from u=1+1=\color{red}\bold 2 to u=2+1=\color{red}\bold 3

    So when you evaluate, we see that

    \int_1^2\frac{\,dx}{(x+1)^2}=\int_2^3\frac{\,du}{u  ^2}=\color{red}\boxed{\frac{1}{6}}

    I hope this clarified things! If you have questions, feel free to ask. We're here to help you!

    --Chris
    I see.. thanks
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by mojo0716 View Post
    The solution manual has it like this, though, and I wonder what they did? Looks like they didn't use any substitution.

    \int_{1}^{2} 1/(x+1)^2 \: dx = (-1/(x+1)) from 1 to 2

    after that I know what to do, but I don't know how they got = (-1/(x+1)) from 1 to 2

    thanks
    That is because they did not use substitution. They integarated it directly.

    INT[1 / (1+x)^2]dx
    = INT[(1 +x)^(-2)]dx
    = 1/(-1) *(1 +x)^(-1) +C
    = -(1 +x)^(-1) +C
    = -1 / (1 +x) +C

    And so the boundaries for the integration for dx remained from x=1 to x=2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. An integral to solve
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 14th 2011, 02:00 AM
  2. Solve Integral with TI-84 Plus
    Posted in the Calculus Forum
    Replies: 9
    Last Post: February 13th 2011, 07:24 PM
  3. Replies: 1
    Last Post: June 9th 2009, 11:37 PM
  4. How to solve this integral?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 4th 2008, 11:52 AM
  5. How do you solve this integral?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 15th 2008, 01:45 AM

Search Tags


/mathhelpforum @mathhelpforum