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Thread: tricky integral

  1. #1
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    tricky integral

    I'm having trouble with this integral:

    $\displaystyle \int{\frac{x}{\sqrt{x^2 + x + 1}}}$

    I don't see an obvious way to use u-substitution, integration by parts, or trig substitution.

    Could someone show me the first few steps?
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  2. #2
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    Quote Originally Posted by centenial View Post
    I'm having trouble with this integral:

    $\displaystyle \int{\frac{x}{\sqrt{x^2 + x + 1}}}$

    I don't see an obvious way to use u-substitution, integration by parts, or trig substitution.

    Could someone show me the first few steps?
    Wolfram shows that it's rather nasty ...

    integrate x/sqrt(x^2+x+1) - Wolfram|Alpha

    click on "show steps"
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  3. #3
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    It's the third time I solve this integral in the forum, where is it coming from?

    Start by rewriting the integral as:

    $\displaystyle \displaystyle I = \int\frac{x}{\sqrt{x^2+x+1}}\;{dx} = \int\frac{\frac{1}{2}(2x+1)+x-\frac{1}{2}(2x+1)}{\sqrt{x^2+x+1}}\;{dx}$

    $\displaystyle \displaystyle \Rightarrow I = \int\frac{\frac{1}{2}(2x+1)}{\sqrt{x^2+x+1}}\;{dx} +\frac{1}{2}\int\frac{2x-(2x+1)}{\sqrt{x^2+x+1}}\;{dx}$

    $\displaystyle \displaystyle \Rightarrow I = \underbrace{\int\frac{\frac{1}{2}(2x+1)}{\sqrt{x^2 +x+1}}\;{dx}}_{I_{1}}-\underbrace{\frac{1}{2}\int\frac{1}{\sqrt{x^2+x+1} }\;{dx}}_{I_{2}} $

    We can see that $\displaystyle I_{1}$ is of the form $\displaystyle \int\frac{\frac{1}{2}f'(x)}{\sqrt{f(x)}}\;{dx} = \sqrt{f(x)}+C$
    (or we can just let $\displaystyle u = \sqrt{x^2+x+1}$). Thus $\displaystyle I_{1} = \sqrt{x^2+x+1}+k_{1}$

    For $\displaystyle I_{2}$ complete the square $\displaystyle x^2+x+1 = \left(x+\frac{1}{2}\right)^2+\frac{3}{4}.$
    Then let $\displaystyle x+\frac{1}{2} = \frac{\sqrt{3}}{2}\sinh{\varphi}$, then $\displaystyle dx = \frac{\sqrt{3}}{2}\cosh{\varphi} \;{d\varphi}$.

    $\displaystyle \displaystyle \therefore ~ I_{2} = \frac{1}{2}\int\frac{1}{\sqrt{x^2+x+1}}\;{dx} = \frac{1}{2}\int\frac{1}{\sqrt{\left(x+\frac{1}{2}\ right)^2+\frac{3}{4}}}\;{dx}$

    $\displaystyle \displaystyle \Rightarrow ~ I_{2} = \frac{1}{2}\int\frac{\frac{\sqrt{3}}{2}\cosh{\varp hi}}{\sqrt{\left(\frac{\sqrt{3}}{2}\sinh{\varphi}\ right)^2+\frac{3}{4}}}\;{d\varphi} = \frac{1}{2}\int\frac{\frac{\sqrt{3}}{2}\cosh{\varp hi}}{\sqrt{\frac{3}{4}\sinh^2{\varphi}+\frac{3}{4} }}\;{d\varphi}$

    $\displaystyle \displaystyle\Rightarrow ~ I_{2} = \frac{1}{2}\int\frac{\frac{\sqrt{3}}{2}\cosh{\varp hi}}{\sqrt{\frac{3}{4}\cosh^2{\varphi}}}\;{d\varph i} = \frac{1}{2}\int\frac{\frac{\sqrt{3}}{2}\cosh{\varp hi}}{\frac{\sqrt{3}}{2}\cosh{\varphi}}\;{d\varphi} $

    $\displaystyle \displaystyle \Rightarrow ~ I_{2} = \frac{1}{2}\int\;{d\varphi} = \frac{1}{2}\varphi+k_{2} = \frac{1}{2}\sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+k_{2} $

    Therefore $\displaystyle \displaystyle I = I_{1}-I_{2} = \sqrt{x^2+x+1}-\frac{1}{2}\sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+k$.
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