How to integrate this u substitution

the integral of cube root of x^3 + 5 times x^5

∫³√(x³+5) *x⁵ dx=
=∫³√(x³+5) *x³ * x² dx=
=(1/3)∫³√(x³+5) *x³ d(x³)=
=(1/3)∫³√(x³+5) *x³ d(x³+5)=
Let v=x³+5
=(1/3)∫³√v *(v-5) dv=
=(1/3)∫ v^(1/3+1) - 5v^(1/3) dv=
=(1/3)∫ v^(4/3) dx - (5/3)∫ v^(1/3) dv=
=(1/3)v^(4/3+1)/(4/3+1) - (5/3)v^(1/3+1)/(1/3+1) +C=
=(1/7)v^(7/3) - (5/4)v^(4/3) +C=
=(1/7)(x³+5)^(7/3) - (5/4)(x³+5)^(4/3) +C

(x^3 + 5x^5)^(1/3) * dx
(x^3 * (1 + 5x^2))^(1/3) * dx
(x^3)^(1/3) * (1 + 5x^2)^(1/3) * dx
x * (1 + 5x^2)^(1/3) * dx

u = 1 + 5x^2
du = 10x * dx

(1/10) * u^(1/3) * du

Integrate

(1/10) * (3/4) * u^(4/3) + C
(3/40) * (1 + 5x^2)^(4/3) + C