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Thread: How do you solve this integral?

  1. #1
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    How do you solve this integral?

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

    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
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mojo0716 View Post
    $\displaystyle \int_{1}^{2} 1/(x+1)^2 \: dx$
    Make the u substitution $\displaystyle u=x+1\implies \,du=\,dx$

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

    I'm sure you can take it from here.

    --Chris
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Make the u substitution $\displaystyle u=x+1\implies \,du=\,dx$

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

    I'm sure you can take it from here.

    --Chris
    ah thanks
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  4. #4
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    The solution manual has it like this, though, and I wonder what they did? Looks like they didn't use any substitution.

    $\displaystyle \int_{1}^{2} 1/(x+1)^2 \: dx$ = (-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
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mojo0716 View Post
    The solution manual has it like this, though, and I wonder what they did?

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



    thanks
    There way is correct, and so is mine.

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

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

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

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

    So when you evaluate, we see that

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

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

    --Chris
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    There way is correct, and so is mine.

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

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

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

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

    So when you evaluate, we see that

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

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

    --Chris
    I see.. thanks
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  7. #7
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    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.

    $\displaystyle \int_{1}^{2} 1/(x+1)^2 \: dx$ = (-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.
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