<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] Factor the denominator and expand the rational function into partial fractions:
\begin{eqnarray*}
\frac{1}{x^{4}-5x^{2}+16} &=&\frac{1}{\left( x^{2}-\sqrt{13}x+4\right)
\left( x^{2}+\sqrt{13}x+4\right) } \\
&=&\frac{1}{104}\frac{\sqrt{13}x+13}{x^{2}+\sqrt{13}x+4}-\frac{1}{104}\frac{
\sqrt{13}x-13}{x^{2}-\sqrt{13}x+4}.
\end{eqnarray*}
Then complete the square in the denominators
\begin{equation*}
\frac{1}{x^{4}-5x^{2}+16}=\frac{\sqrt{13}}{104}\frac{x+13/\sqrt{13}}{\left(
x+\sqrt{13}/2\right) ^{2}+3/4}-\frac{\sqrt{13}}{104}\frac{x-13/\sqrt{13}}{
\left( x-\sqrt{13}/2\right) ^{2}+3/4},
\end{equation*}
and use, respectivelly,  the substitutions $$u=x+\sqrt{13}/2,\qquad \text{and}\qquad v=x-\sqrt{13}/2.$$
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