This integral is solved by making the substitution u = 2x3 + 7 and using the power rule. We find du = 6x2 dx. After the substitution, we have ∫ u–3/2 du, which is solved using the power rule with n = –3/2.

u = 2x3 + 7;   du = 6x2 dx
∫   6x2 dx
(2x3 + 7)3/2
=   ∫   du
u3/2
=  ∫ u–3/2 du
=  –2u–1/2  + C
=  –2 (2x3 + 7)–1/2  + C