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Math Help - Integral containing something over the root of x-a

  1. #1
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    Integral containing something over the root of x-a

    Hi there, I'm facing this problem which I couldn't solve.

    The statement says:
    Evaluate the next integrals, expressing it previously in the forms that contains \sqrt[ ]{a^2+u^2}, \sqrt[ ]{a^2-u^2} and \sqrt[ ]{u^2-a^2}, and then solve it using the integral table;

    The exercise which I couldn't solve is:

    \displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sq  rt[ ]{x^2+2x}}dx

    I started by completing the square:

    x^2+2x=(x+1)^2-1

    Then

    u^2=(x+1)^2
    u=(x+1)\Rightarrow{x+3=u+2}
    du=dx

    \displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sq  rt[ ]{x^2+2x}}dx=\displaystyle\int_{}^{}\displaystyle\f  rac{u+2}{\sqrt[ ]{u^2-1}}du

    From here I've tried to solve it by parts
    t=u+2
    dt=du

    dv=\displaystyle\frac{1}{\sqrt[ ]{u^2-1}}du
    v=\ln|u+\sqrt[ ]{u^2-1}|

    \displaystyle\int_{}^{}\displaystyle\frac{u+2}{\sq  rt[ ]{u^2-1}}du=(u+2)\ln|u+\sqrt[ ]{u^2-1}|-\displaystyle\int_{}^{}\ln|u+\sqrt[ ]{u^2-1}|du

    If I try again by parts with -\displaystyle\int_{}^{}\ln|u+\sqrt[ ]{u^2-1}|du it gets more complicated.

    So, what do you say?

    Bye there.
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    Ok, I think I have solved it, here is the solution I have arrived to:

    \displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sq  rt[ ]{x^2+2x}}dx=\sqrt[ ]{(x+1)^2-1}+2\ln|x+1+\sqrt[ ]{(x+1)^2-1}|+C
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  5. #5
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    Quote Originally Posted by Ulysses View Post
    Ok, I think I have solved it, here is the solution I have arrived to:

    \displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sq  rt[ ]{x^2+2x}}dx=\sqrt[ ]{(x+1)^2-1}+2\ln|x+1+\sqrt[ ]{(x+1)^2-1}|+C
    That's correct!

    Here is how I went about it: \displaystyle \int\frac{x+3}{\sqrt{(x^2+2x)}}\;{dx} = \int\frac{x+3}{\sqrt{(x+1)^2-1}}\;{dx}. Let x+1 = \cosh{\theta}, then \dfrac{dx}{d\theta} = \sinh{\theta} \Rightarrow {dx} = \sinh{\theta}\;{d\theta}. Also x+1 = \cosh{\theta} \Leftrightarrow x+3 = \cosh{\theta}+2. Putting this together we have \displaystyle \int\frac{x+3}{\sqrt{(x^2+2x)}}\;{dx} = \int\frac{(\cosh{\theta}+2)(\sinh{\theta})}{\sqrt{  {\cosh^2{\theta}}-1}}\;{d\theta}  \displaystyle = \int\frac{(\cosh{\theta}+2)(\sinh{\theta})}{\sqrt{  \sinh^2{\theta}}}\;{d\theta} = \int\frac{(\cosh{\theta}+2)(\sinh{\theta})}{(\sinh  {\theta})}\;{d\theta} \displaystyle = \int\cosh{\theta}+2\;{d\theta} = \sinh{\theta}+2{\theta}+k, where \theta = \text{arccosh}(x+1) of course. Using the little beautiful identity \sinh\left\{\text{arccosh}\varphi\right\} = \sqrt{\varphi-1}\sqrt{\varphi+1} this reduces to \sqrt{x}\sqrt{x+2}+2\text{arccosh}(x+1)+k.
    Last edited by TheCoffeeMachine; June 30th 2010 at 06:01 PM.
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