Answer
$$\frac{{{{\left( {{x^4} + 1} \right)}^{3/2}}}}{{30}}\left( {3{x^4} - 2} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {{x^7}\sqrt {{x^4} + 1} } dx \cr
& {\text{integrate by the substitution method}} \cr
& {\text{set }}u = {x^4} + 1,\,\,\,\,{x^4} = u - 1{\text{ then }}du = 4{x^3}dx,\,\,\,\,dx = \frac{{du}}{{4{x^3}}} \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{x^7}\sqrt {{x^4} + 1} } dx = \int {{x^7}\sqrt u } \left( {\frac{{du}}{{4{x^3}}}} \right) \cr
& = \int {\left( {u - 1} \right)\sqrt u } \left( {\frac{{du}}{4}} \right) \cr
& = \frac{1}{4}\int {\left( {u - 1} \right){u^{1/2}}} du \cr
& = \frac{1}{4}\int {\left( {{u^{3/2}} - {u^{1/2}}} \right)} du \cr
& {\text{integrating}}{\text{,}} \cr
& = \frac{1}{4}\left( {\frac{{{u^{5/2}}}}{{5/2}} - \frac{{{u^{3/2}}}}{{3/2}}} \right) + C \cr
& = \frac{2}{{4\left( {15} \right)}}\left( {3{u^{5/2}} - 5{u^{3/2}}} \right) + C \cr
& = \frac{{{u^{3/2}}}}{{30}}\left( {3u - 5} \right) + C \cr
& {\text{replace }}{x^4} + 1{\text{ for }}u \cr
& = \frac{{{{\left( {{x^4} + 1} \right)}^{3/2}}}}{{30}}\left( {3\left( {{x^4} + 1} \right) - 5} \right) + C \cr
& = \frac{{{{\left( {{x^4} + 1} \right)}^{3/2}}}}{{30}}\left( {3{x^4} - 2} \right) + C \cr} $$
