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Thread: 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 $\displaystyle \sqrt[ ]{a^2+u^2}$, $\displaystyle \sqrt[ ]{a^2-u^2}$ and $\displaystyle \sqrt[ ]{u^2-a^2}$, and then solve it using the integral table;

    The exercise which I couldn't solve is:

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

    I started by completing the square:

    $\displaystyle x^2+2x=(x+1)^2-1$

    Then

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

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

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

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

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

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

    So, what do you say?

    Bye there.
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  3. #3
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  4. #4
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    Ok, I think I have solved it, here is the solution I have arrived to:

    $\displaystyle \displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sq rt[ ]{x^2+2x}}dx=\sqrt[ ]{(x+1)^2-1}+2\ln|x+1+\sqrt[ ]{(x+1)^2-1}|+C$
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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:

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

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