Answer
$$\frac{{{{\left( {{x^4} + 2} \right)}^{3/2}}}}{6} - {\left( {{x^4} + 2} \right)^{1/2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^7}}}{{\sqrt {{x^4} + 2} }}} dx \cr
& {\text{Let }}u = {x^4} + 2{\text{ }} \Rightarrow {\text{ }}du = 4{x^3}dx \cr
& \int {\frac{{{x^7}}}{{\sqrt {{x^4} + 2} }}} dx = \int {\frac{{{x^4}}}{{\sqrt {{x^4} + 2} }}} \left( {{x^3}} \right)dx = \int {\frac{{u - 2}}{{{u^{1/2}}}}} \left( {\frac{1}{4}} \right)du \cr
& {\text{ }} = \frac{1}{4}\int {\left( {{u^{1/2}} - 2{u^{ - 1/2}}} \right)} du \cr
& {\text{ }} = \frac{1}{4}\left( {\frac{{{u^{3/2}}}}{{3/2}} - \frac{{2{u^{1/2}}}}{{1/2}}} \right) + C \cr
& {\text{ }} = \frac{1}{4}\left( {\frac{{2{u^{3/2}}}}{3} - 4{u^{1/2}}} \right) + C \cr
& {\text{ }} = \frac{{{u^{3/2}}}}{6} - {u^{1/2}} + C \cr
& {\text{Back - substitute }}u = {x^4} + 2 \cr
& {\text{ }} = \frac{{{{\left( {{x^4} + 2} \right)}^{3/2}}}}{6} - {\left( {{x^4} + 2} \right)^{1/2}} + C \cr} $$
