Question:

Let $I = \frac{1}{2} \int \frac{2}{x^4 + 5x^2 + 1} dx$<br/>$\Rightarrow I = \frac{1}{2} \int \frac{1 + x^2}{x^4 + 5x^2 + 1} dx + \frac{1}{2} \int \frac{1 - x^2}{x^4 + 5x^2 + 1} dx = \frac{1}{2} \int \frac{1 + \frac{1}{x^2}}{x^2 + 5 + \frac{1}{x^2}} dx - \frac{1}{2} \int \frac{1 - \frac{1}{x^2}}{x^2 + 5 + \frac{1}{x^2}} dx$<br/>{dividing $N^r$ and $D^r$ by $x^2$}<br/>$= \frac{1}{2} \int \frac{\left(1 + \frac{1}{x^2}\right)}{\left(x - \frac{1}{x}\right)^2 + 7} dx - \frac{1}{2} \int \frac{\left(1 - \frac{1}{x^2}\right)}{\left(x + \frac{1}{x}\right)^2 + 3} dx = \frac{1}{2} \int \frac{dt}{t^2 + (\sqrt{7})^2} - \frac{1}{2} \int \frac{du}{u^2 + (\sqrt{3})^2}$<br/>where $t = x - \frac{1}{x}$ and $u = x + \frac{1}{x}$<br/>$I = \frac{1}{2} \cdot \frac{1}{\sqrt{7}} \left( \tan^{-1} \frac{t}{\sqrt{7}} \right) - \frac{1}{2} \cdot \frac{1}{\sqrt{3}} \left( \tan^{-1} \frac{u}{\sqrt{3}} \right) + C$<br/>$= \frac{1}{2} \left[ \frac{1}{\sqrt{7}} \tan^{-1} \left( \frac{x - \frac{1}{x}}{\sqrt{7}} \right) - \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x + \frac{1}{x}}{\sqrt{3}} \right) \right] + C$ Ans.