Answer
$$\frac{1}{{16}}\ln \left| {\frac{{4{x^2} - 1}}{{4{x^2} + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{16{x^4} - 1}}dx} \cr
& {\text{write the integrand as}} \cr
& = \int {\frac{x}{{{{\left( {4{x^2}} \right)}^2} - 1}}dx} \cr
& {\text{Make an appropiate }}u{\text{ - substitution }} \cr
& u = 4{x^2},\,\,\,\,\,\,\,du = 8xdx,\,\,\,\,\,\,dx = \frac{{du}}{{8x}} \cr
& {\text{write in terms of }}u \cr
& \int {\frac{x}{{{{\left( {4{x^2}} \right)}^2} - 1}}dx} = \int {\frac{x}{{{u^2} - 1}}\left( {\frac{{du}}{{8x}}} \right)} \cr
& = \frac{1}{8}\int {\frac{1}{{{u^2} - 1}}du} \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 70}} \cr
& \left( {70} \right):\,\,\,\,\,\,\int {\frac{{du}}{{{u^2} - {a^2}}} = \frac{1}{{2a}}\ln \left| {\frac{{u - a}}{{u + a}}} \right| + C} \cr
& {\text{take }}a = 1,\,\, \cr
& \frac{1}{8}\int {\frac{1}{{{u^2} - 1}}du} = \frac{1}{{16\left( 1 \right)}}\ln \left| {\frac{{u - 1}}{{u + 1}}} \right| + C \cr
& {\text{simplifying}} \cr
& = \frac{1}{{16}}\ln \left| {\frac{{u - 1}}{{u + 1}}} \right| + C \cr
& {\text{write in terms of }}x{\text{; replace }}4{x^2}{\text{ for }}u \cr
& = \frac{1}{{16}}\ln \left| {\frac{{4{x^2} - 1}}{{4{x^2} + 1}}} \right| + C \cr} $$
