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
 A: 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
A: 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 $$u=x+\sqrt{13}/2,\qquad \text{and}\qquad v=x-\sqrt{13}/2.$$
A: 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.
