<s>[INST] How to integrate this fraction: $\int\frac{1}{1-2x^2}dx$? I'm not sure how to integrate this:
$$\int\frac{1}{1-2x^2}dx$$
I think it has to be this:
$$ -2\cdot \arctan(x)$$
Or this:
$$\arctan(\sqrt{-2x^2})$$
 [/INST] $$ \int \frac{1}{1-2z^2} = \frac{1}{2} \int \frac{1}{1-\sqrt{2}z} + \frac{1}{1+\sqrt{2}z} = \frac{1}{2} (-\frac{1}{\sqrt{2}} \log|1-\sqrt{2}z| + \frac{1}{\sqrt{2}} \log|1+\sqrt{2}z|) = \frac{1}{2\sqrt{2}} \log|\frac{1+\sqrt{2}z}{1-\sqrt{2}z}| $$
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