Answer
$$\frac{1}{{2048}}\ln \left| {\frac{{{t^8} - 256}}{{{t^8}}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dt}}{{t\left( {{t^8} - 256} \right)}}} \cr
& {\text{Let }}x = {t^4},\,\,\,\,dx = 4{t^3}dt \cr
& \cr
& {\text{Use the change of variable}} \cr
& \int {\frac{1}{{t\left( {{t^8} - 256} \right)}}dt} = \int {\frac{1}{{t\left( {{x^2} - 256} \right)}}} \left( {\frac{{dx}}{{4{t^3}}}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4}\int {\frac{1}{{{t^4}\left( {{x^2} - 256} \right)}}} dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4}\int {\frac{1}{{x\left( {{x^2} - {{16}^2}} \right)}}} dx \cr
& {\text{Use the table of integrals formula 75: }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int {\frac{{dx}}{{x\left( {{x^2} - {a^2}} \right)}} = \frac{1}{{2{a^2}}}\ln \left| {\frac{{{x^2} - {a^2}}}{{{x^2}}}} \right| + C} \cr
& {\text{Therefore,}} \cr
& \frac{1}{4}\int {\frac{1}{{x\left( {{x^2} - {{16}^2}} \right)}}} dx = \frac{1}{4}\left( {\frac{1}{{2{{\left( {16} \right)}^2}}}\ln \left| {\frac{{{x^2} - {{16}^2}}}{{{x^2}}}} \right|} \right) + C \cr
& {\text{Simplify}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{2048}}\ln \left| {\frac{{{x^2} - 256}}{{{x^2}}}} \right| + C \cr
& {\text{Substitute }}{t^4}{\text{ for }}x \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{2048}}\ln \left| {\frac{{{{\left( {{t^4}} \right)}^2} - 256}}{{{{\left( {{t^4}} \right)}^2}}}} \right| + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{2048}}\ln \left| {\frac{{{t^8} - 256}}{{{t^8}}}} \right| + C \cr} $$
