I assume you're asked to compute
[tex]\displaystyle \int \frac{30x^2}{\sqrt{x-4}} \, dx[/tex]
using both of the substitutions provided.
With [tex]u=x-4[/tex], we have [tex]x=u+4[/tex] and [tex]dx=du[/tex]. Then
[tex]\displaystyle \int \frac{30x^2}{\sqrt{x-4}} \, dx = \int \frac{30(u+4)^2}{\sqrt u} \, du \\\\ ~~~~~~~~ = 30 \int \frac{u^2 + 8u + 16}{\sqrt u} \, du \\\\ ~~~~~~~~ = 30 \int \left(u^{3/2} + 8u^{1/2} + 16u^{-1/2}\right) \, du \\\\ ~~~~~~~~ = 30 \left(\frac25 u^{5/2} + \frac{16}3 u^{3/2} + 32 u^{1/2}\right) + C \\\\ ~~~~~~~~ = 12 u^{5/2} + 160 u^{3/2} + 960 u^{1/2} + C \\\\ ~~~~~~~~ = 12 (x-4)^{5/2} + 160 (x-4)^{3/2} + 960 (x-4)^{1/2} + C \\\\ ~~~~~~~~ = 4 \sqrt{x-4} \left(3 (x-4)^2 + 40 (x-4) + 240\right) + C \\\\ ~~~~~~~~ = \boxed{4 \sqrt{x-4} \left(3x^2 + 16x + 128\right) + C}[/tex]
With [tex]u=\sqrt{x-4}[/tex], we have
[tex]u^2 = x-4 \implies x^2 = (u^2+4)^2[/tex]
and [tex]2u\,du=dx[/tex]. Then
[tex]\displaystyle \int \frac{30x^2}{\sqrt{x-4}} \, dx = \int \frac{60u \left(u^2+4\right)^2}u \, du \\\\ ~~~~~~~~ = 60 \int \left(u^4 + 8u^2 + 16\right) \, du \\\\ ~~~~~~~~ = 60 \left(\frac15 u^5 + \frac83 u^3 + 16u\right) + C \\\\ ~~~~~~~~ = 12 (x-4)^{5/2} + 160 (x-4)^{3/2} + 960 (x-4)^{1/2} + C \\\\ ~~~~~~~~ = 4 \sqrt{x-4} \left(3 (x-4)^2 + 40 (x-4) + 240\right) + C \\\\ ~~~~~~~~ = \boxed{4\sqrt{x-4} \left(3x^2 + 16x + 128\right) + C}[/tex]
