juantheron
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Re: MX2 2016 Integration Marathon
![](https://latex.codecogs.com/png.latex?\bg_white $let $I = \int\frac{1}{\left(x+\sqrt{x^2-1}\right)^{2}}dx$\;, Now put $(x+\sqrt{x^2-1}) = t\;,$)
![](https://latex.codecogs.com/png.latex?\bg_white $Then $(x+\sqrt{x^2-1}) dx = \sqrt{x^2-1}dt$)
![](https://latex.codecogs.com/png.latex?\bg_white $Now Using $(x+\sqrt{x^2-1})\cdot (x-\sqrt{x^2-1})=1 \;,$ Then $(x-\sqrt{x^2-1}) =\frac{1}{t}$)
![](https://latex.codecogs.com/png.latex?\bg_white $So we get $\sqrt{x^2-1} = \frac{1}{2}\left(t-\frac{1}{t}\right) = \frac{t^2-1}{2t}$)
![](https://latex.codecogs.com/png.latex?\bg_white $ So $I = \frac{1}{2}\int \frac{t^2-1}{t^3}dt = \frac{1}{2}\ln |t|+\frac{1}{4}t^2+\mathcal{C}$)
![](https://latex.codecogs.com/png.latex?\bg_white $ So $I = \frac{1}{2}\ln \left|x+\sqrt{x^2-1}\right|+\frac{1}{4}\cdot \frac{1}{\left(x+\sqrt{x^2-1}\right)^2}+\mathcal{C}$)
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.