Answer
$$\frac{1}{{80}}\ln \left| {\frac{{{x^5} + 4}}{{{x^5}}}} \right| - \frac{1}{{20{x^5}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^6}\left( {{x^5} + 4} \right)}}} dx \cr
& {\text{Multiply by }}\frac{{{x^4}}}{{{x^4}}} \cr
& = \int {\frac{{{x^4}}}{{{x^{10}}\left( {{x^5} + 4} \right)}}} dx \cr
& {\text{Let }}{x^5} = u,\,\,\,\,5{x^4}dx = du \cr
& = \int {\frac{{\left( {1/5} \right)du}}{{{u^2}\left( {u + 4} \right)}}} \cr
& = \frac{1}{5}\int {\frac{{du}}{{{u^2}\left( {u + 4} \right)}}} \cr
& {\text{Decompose }}\frac{1}{{{u^2}\left( {u + 4} \right)}}{\text{ into partial fractions}} \cr
& \frac{1}{{{u^2}\left( {u + 4} \right)}} = \frac{A}{u} + \frac{B}{{{u^2}}} + \frac{C}{{u + 4}} \cr
& {\text{Multiply by }}u\left( {u + 1} \right){\text{ and simplify}} \cr
& 1 = Au\left( {u + 4} \right) + B\left( {u + 4} \right) + C{u^2} \cr
& 1 = A{u^2} + 4Au + Bu + 4B + C{u^2} \cr
& 1 = \left( {A{u^2} + C{u^2}} \right) + 4Au + Bu + 4B \cr
& {\text{Equate the coefficients}} \cr
& A + C = 0,\,\,\,\,4A + B = 0,\,\,\,4B = 1 \cr
& {\text{Solve to get}} \cr
& B = 1/4,\,\,\,\,A = - 1/16,\,\,\,\,\,C = 1/16 \cr
& \cr
& {\text{Replace the coefficients}} \cr
& \frac{1}{{{u^2}\left( {u + 4} \right)}} = \frac{{ - 1/16}}{u} + \frac{{1/4}}{{{u^2}}} + \frac{{1/16}}{{u + 4}} \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& \frac{1}{5}\int {\frac{{du}}{{{u^2}\left( {u + 4} \right)}}} = \frac{1}{5}\int {\left( {\frac{{ - 1/16}}{u} + \frac{{1/4}}{{{u^2}}} + \frac{{1/16}}{{u + 4}}} \right)} du \cr
& {\text{Integrate}} \cr
& = \frac{1}{5}\left( { - \frac{1}{{16}}\ln \left| u \right| - \frac{1}{{4u}} + \frac{1}{{16}}\ln \left| {u + 4} \right|} \right) + C \cr
& = - \frac{1}{{80}}\ln \left| u \right| - \frac{1}{{20u}} + \frac{1}{{80}}\ln \left| {u + 4} \right| + C \cr
& \cr
& {\text{Replace }}u = {x^5} \cr
& = - \frac{1}{{80}}\ln \left| {{x^5}} \right| - \frac{1}{{20{x^5}}} + \frac{1}{{80}}\ln \left| {{x^5} + 4} \right| + C \cr
& = \frac{1}{{80}}\ln \left| {\frac{{{x^5} + 4}}{{{x^5}}}} \right| - \frac{1}{{20{x^5}}} + C \cr} $$
