how to find the definite integral of (x^3)/((x^2)+4)^0.5 of x from 0 to 2 ?

∫[0 ~ 2] x³/√(x² + 4) dx

Let y = x² + 4, then dy = 2x dx.
When x = 0, y = 4; when x = 2, y = 8.

∫[0 ~ 2] x³/√(x² + 4) dx
= ∫[4 ~ 8] (1/2) (y - 4)/√y dy
= (1/2) ∫[4 ~ 8] (√y - 4/√y) dy
= (1/2) [ (2/3)y^(3/2) - 8√y ]@(8 and 4)
= (1/2) [ (16/3)√8 - 8√8 - 16/3 + 16 ]
= (1/2) [ (-8/3)√8 + 32/3 ]
= (1/2) [ (-16/3)√2 + 32/3 ]
= (-8/3)√2 + 16/3
= 16/3 - (8√2)/3

∫x^3/√(x² + 4) dx x from 0 to 2
=(1/3)∫d(x^3)/√x^3 x from 0 to 2
=(2/15)x^(5/2) |x from 0 to 2
=2.2627417