Answer
$\frac{1}{5}\ln \left| {{x^5} + \sqrt {{x^{10}} - 2} } \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^4}}}{{\sqrt {{x^{10}} - 2} }}} dx \cr
& {\text{Let }}u = {x^5},{\text{ }}du = 5{x^4}dx,{\text{ }}{x^4}dx = \frac{1}{5}du{\text{, substituting we obtain}} \cr
& \int {\frac{{{x^4}}}{{\sqrt {{x^{10}} - 2} }}} dx = \int {\frac{{\left( {1/5} \right)}}{{\sqrt {{u^2} - 2} }}} du \cr
& = \frac{1}{5}\int {\frac{1}{{\sqrt {{u^2} - 2} }}} du \cr
& {\text{Using the entry 43 }}\left( {{\text{Integrals on Reference Pages 6}}--{\text{1}}0} \right) \cr
& \int {\frac{{du}}{{\sqrt {{u^2} - {a^2}} }}} = \ln \left| {u + \sqrt {{u^2} - {a^2}} } \right| + C,{\text{ therefore}} \cr
& \frac{1}{5}\int {\frac{1}{{\sqrt {{u^2} - 2} }}} du = \frac{1}{5}\ln \left| {u + \sqrt {{u^2} - 2} } \right| + C \cr
& {\text{Write in terms of }}u,{\text{ }}u = {x^5},{\text{ then}} \cr
& \int {\frac{{{x^4}}}{{\sqrt {{x^{10}} - 2} }}} dx = \frac{1}{5}\ln \left| {{x^5} + \sqrt {{x^{10}} - 2} } \right| + C \cr} $$
