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  1. MHB Master
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    #1
    7.4.32 Evaluate the integral
    $\displaystyle \int_0^1\dfrac{x}{x^2+4x+13}\, dx$
    ok side work to complete the square
    $x^2+4x=-13$
    add 4 to both sides
    $x^2+4x+4=-13+4$
    simplify
    $(x+2)^2+9=0$
    ok now what


    W|A returned ≈0.03111
    Last edited by karush; October 7th, 2019 at 16:37.

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    #2
    Quote Originally Posted by karush View Post
    7.4.32 Evaluate the integral
    $\displaystyle \int_0^1\dfrac{x}{x^2+4x+13}\, dx$
    ok side work to complete the square
    $x^2+4x=-13$
    add 4 to both sides
    $x^2+4x+4=-13+4$
    simplify
    $(x+2)^2+9=0$
    ok now what
    $u = x+2 \implies du = dx$

    $ \displaystyle \int_0^1 \dfrac{(x+2)-2}{(x+2)^2 + 9} \, dx$

    substitute and reset limits of integration ...

    $ \displaystyle \int_2^3 \dfrac{u}{u^2 + 9} - \dfrac{2}{u^2+9} \, du$

    looks like the result will be a log and an inverse tangent ...

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    #3 Thread Author
    ok little ??? about where $u=x+2$ and the limits work...

    $$\Biggr|\dfrac{1}{2} \ln(u^2 + 9)-\frac{2}{3}\arctan \left(\frac{u}{3}\right)\Biggr|_2^3$$

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    #4
    Quote Originally Posted by karush View Post
    ok little ??? about where $u=x+2$ and the limits work...

    $\Bigg[\dfrac{1}{2} \ln(u^2 + 9)-\frac{2}{3}\arctan \left(\frac{u}{3}\right)\Bigg]_2^3$
    $u=x+2$

    lower limit of integration is $x=0 \implies x+2 = 0+2 = 2$, the lower limit of integration reset to a $u$ value

    do the same for the upper limit of integration

    $\Bigg[\dfrac{1}{2} \ln(u^2 + 9)-\dfrac{2}{3}\arctan \left(\dfrac{u}{3}\right)\Bigg]_2^3$

    $\left[\dfrac{1}{2}\ln(18)-\dfrac{2}{3}\arctan(1)\right] - \left[\dfrac{1}{2}\ln(13) - \dfrac{2}{3}\arctan\left(\dfrac{2}{3}\right)\right]$

    $\ln{\sqrt{\dfrac{18}{13}}} - \dfrac{\pi}{6} + \dfrac{2}{3}\arctan\left(\dfrac{2}{3} \right) \approx 0.03111$

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