Answer
$$\frac{1}{2}\sqrt {64 - {x^4}} - 4\ln \left| {\frac{{8 + \sqrt {64 - {x^4}} }}{{{x^2}}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {64 - {x^4}} }}{x}} dx \cr
& {\text{Use a table of integrals with forms involving }}\sqrt {{a^2} - {u^2}} \cr
& {\text{From the table of integrals in the back of the book}} \cr
& {\text{Formula 39}} \cr
& \int {\frac{{\sqrt {{a^2} - {u^2}} }}{u}du = \sqrt {{a^2} - {u^2}} - a\ln \left| {\frac{{a + \sqrt {{a^2} - {u^2}} }}{u}} \right| + C} \cr
& \int {\frac{{\sqrt {64 - {x^4}} }}{x}} dx{\text{ can be written as }}\int {\frac{{\sqrt {{{\left( 8 \right)}^2} - {{\left( {{x^2}} \right)}^2}} }}{x}} dx \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx,{\text{ }}dx = \frac{{du}}{{2x}} \cr
& \int {\frac{{\sqrt {{{\left( 8 \right)}^2} - {{\left( {{x^2}} \right)}^2}} }}{x}} dx = \int {\frac{{\sqrt {{{\left( 8 \right)}^2} - {u^2}} }}{x}} \frac{{du}}{{2x}} = \int {\frac{{\sqrt {{{\left( 8 \right)}^2} - {u^2}} }}{{2{x^2}}}} du \cr
& = \frac{1}{2}\int {\frac{{\sqrt {{{\left( 8 \right)}^2} - {u^2}} }}{{{u^2}}}} du \to u = {x^2},{\text{ }}a = 8 \cr
& {\text{Substituting into formula 39}} \cr
& = \frac{1}{2}\left( {\sqrt {{{\left( 8 \right)}^2} - {{\left( {{x^2}} \right)}^2}} - 8\ln \left| {\frac{{8 + \sqrt {{{\left( 8 \right)}^2} - {{\left( {{x^2}} \right)}^2}} }}{{{x^2}}}} \right|} \right) + C \cr
& {\text{Multiply and simplify}} \cr
& = \frac{1}{2}\left( {\sqrt {64 - {x^4}} - 8\ln \left| {\frac{{8 + \sqrt {64 - {x^4}} }}{{{x^2}}}} \right|} \right) + C \cr
& = \frac{1}{2}\sqrt {64 - {x^4}} - 4\ln \left| {\frac{{8 + \sqrt {64 - {x^4}} }}{{{x^2}}}} \right| + C \cr} $$
