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  1. #1
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    tricky integral

    I'm having trouble with this integral:

    \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:

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

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

    \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 \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\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 \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 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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