<s>[INST] Evaluation of $\int\frac{1}{x^4-5x^2+16}dx$ 
Evaluation of $\displaystyle \int\frac{1}{x^4-5x^2+16}\,dx$

$\bf{My\; Try::}$ Given $$\displaystyle \int\frac{1}{x^4-5x^2+16}dx = \frac{1}{8}\int\frac{\left(x^2+4\right)-\left(x^2-4\right)}{x^4-5x^2+16}\,dx$$
So We get $$\displaystyle = \frac{1}{8}\int\frac{x^2+4}{x^4-5x^2+16}\,dx-\frac{1}{8}\int\frac{x^2-4}{x^4-5x^2+16}\,dx$$
So We get  $$\displaystyle = \frac{1}{8}\int\frac{1+\frac{4}{x^2}}{\left(x-\frac{4}{x}\right)^2+\left(\sqrt{3}\right)^2}dx-\frac{1}{8}\int\frac{1-\frac{4}{x^2}}{\left(x+\frac{4}{x}\right)^2-\left(\sqrt{13}\right)^2}\,dx$$
Now Using $$\displaystyle \left(x-\frac{4}{x}\right)=t$$ and $$\displaystyle \left(1+\frac{4}{x^2}\right)dx = dt$$ in First Integral and 
Using $$\displaystyle \left(x+\frac{4}{x}\right)=u$$ and $$\displaystyle \left(1-\frac{4}{x^2}\right)\,dx = du$$ in Second Integral.
So We Get $$\displaystyle = \frac{1}{8}\int\frac{1}{t^2+\left(\sqrt{3}\right)^2}dt-\frac{1}{8}\int\frac{1}{u^2-\left(\sqrt{13}\right)^2}\,du$$
So We Get $$\displaystyle = \frac{1}{8\sqrt{3}}\tan^{-1}\left(\frac{x^2-4}{\sqrt{3}x}\right)-\frac{1}{16\sqrt{13}}\ln \left|\frac{x^2-\sqrt{13}x+4}{x^2+\sqrt{13}x+4}\right|+{C}$$
My question is can we solve the above question any other Method, If yes then
please explain here. Thanks
 [/INST] Use hyperbolic functions. The integrand can be stated as $ \frac {4}{4x^4 - 20x^2 + 64} = \frac {4}{(2x^2 - 5)^2 + 39} $
Let $ x = \sqrt{ \frac{5}{2}  }\cosh y $, then $ \mathrm{d} x = \sqrt{ \frac {5}{2} } \sinh y \space \mathrm{d} y $
The integral becomes
$ \displaystyle 4 \cdot \sqrt{ \frac {5}{2} } \int \frac {\sinh y}{5 \sinh^2 y + 39 } \space \mathrm{d} y $
Using the identity $ \sinh^2 A - \cosh^2 A = -1 $, apply another substitution $ \cosh y = \frac {1}{\sqrt 5} z $
Then apply $ \int \frac {1}{x^2+a^2} \mathrm{d}x = \frac {1}{a} \ \arctan \left ( \frac {x}{a} \right ) $ for $ x > 0 $
Back substitute everything and you get your answer
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