Answer
$\frac{1}{{32}}\ln \left| {\frac{{x - 2}}{{x + 2}}} \right| - \frac{1}{{16}}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^4} - 16}}} dx \cr
& {\text{Factor the denominator}} \cr
& = \int {\frac{1}{{\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)}}} dx \cr
& {\text{Decomposing the integrand into partial fractions}} \cr
& \frac{1}{{\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)}} = \frac{{Ax + B}}{{{x^2} - 4}} + \frac{{Cx + D}}{{{x^2} + 4}} \cr
& 1 = \left( {Ax + B} \right)\left( {{x^2} + 4} \right) + \left( {Cx + D} \right)\left( {{x^2} - 4} \right) \cr
& 1 = A{x^3} + 4Ax + B{x^2} + 4B + C{x^3} - 4Cx + D{x^2} - 4D \cr
& 1 = \left( {A{x^3} + C{x^3}} \right) + \left( {B{x^2} + D{x^2}} \right) + 4Ax - 4Cx + 4B - 4D \cr
& A + C = 0 \cr
& B + D = 0 \cr
& 4A - 4C = 0 \cr
& 4B - 4D = 1 \cr
& {\text{Solving the system of equations, we obtain}} \cr
& A = 0,{\text{ }}B = \frac{1}{8},{\text{ }}C = 0,{\text{ }}D = - \frac{1}{8} \cr
& \frac{1}{{\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)}} = \frac{{\frac{1}{8}}}{{{x^2} - 4}} - \frac{{\frac{1}{8}}}{{{x^2} + 4}} \cr
& \int {\frac{1}{{\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)}}} dx = \frac{1}{8}\int {\frac{1}{{{x^2} - 4}}} dx - \frac{1}{8}\int {\frac{1}{{{x^2} + 4}}} dx \cr
& {\text{Integrating by tables on the page 503}} \cr
& = \frac{1}{8}\left( {\frac{1}{{2\left( 2 \right)}}\ln \left| {\frac{{x - 2}}{{x + 2}}} \right|} \right) - \frac{1}{8}\left( {\frac{1}{{2}}{{\tan }^{ - 1}}\left( {\frac{x}{2}} \right)} \right) + C \cr
& = \frac{1}{{32}}\ln \left| {\frac{{x - 2}}{{x + 2}}} \right| - \frac{1}{{16}}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr} $$
