Question:

Let $I = \int \frac{1 - x^2}{1 + x^4} dx = \int \frac{\frac{1}{x^2} - 1}{x^2 + \frac{1}{x^2}} dx$<br/>Put $x + \frac{1}{x} = t \Rightarrow x^2 + \frac{1}{x^2} = t^2 - 2$ and $\left( 1 - \frac{1}{x^2} \right) dx = dt$<br/>$\therefore I = \int \frac{-dt}{t^2 - 2} = \frac{1}{2\sqrt{2}} \ln \left| \frac{t + \sqrt{2}}{t - \sqrt{2}} \right| + C$<br/>$= \frac{1}{2\sqrt{2}} \ln \left| \frac{x^2 + \sqrt{2}x + 1}{x^2 - \sqrt{2}x + 1} \right| + C$