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

You could notice that $x^4-5x^2+16$ can be factored as $(x-a)(x-b)(x+a)(x+b)$ where $a$ and $b$ are complex numbers. Then partial fraction decomposition would lead to quite simple terms and the result for integral will just be the sum of four logarithms.
However, it is more likely that the recombination of the results in terms of reals could be quite tedious.
