juantheron
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Re: MX2 2016 Integration Marathon
^{2}}dx$\;, Now put $(x+\sqrt{x^2-1}) = t\;,$)
 dx = \sqrt{x^2-1}dt$)
\cdot (x-\sqrt{x^2-1})=1 \;,$ Then $(x-\sqrt{x^2-1}) =\frac{1}{t}$)
 = \frac{t^2-1}{2t}$)

^2}+\mathcal{C}$)
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integral of (cos x ) / (2 - cos x )
Your solution to your problem isn't correct, and it's not the same as what Drsoccerball posted.
You should learn Latexintegrate 1/(x^2+2x-1)^(1/2)
I have a bad feeling that (x+1)=tan(theta) is now required.