using the substitution u=6-x^2 integrate (x^3)/(6-x^2)^1/2 with respect to x, between 1 and 2

First we calculate du/dx = -2x and rearrange to get dx as the subject, dx=du/(-2x). Now we change the limits of integration because we are now integrating with respect to u. So the bottom limit will be u(1) = 6-1^2=5 and the top will be u(2) = 6- 2^2 = 2. Now subbing these into the integral we get ∫ -x^2/(2(6-x^2)^1/2) du between 5 and 2 then sub in u to attain ∫ (u-6)/(2u^1/2) du = ∫ (1/2)(u^1/2 - 6u^-1/2)du = [(1/3)u^3/2 -6u^1/2]between 5 and 2 so the integral equals -16/3(2)^1/2 -(5/3(5^1/2)-6(5^1/2)) = 13/3(5^1/2) - 16/3(2^1/2).