What is the integral ofintegral of (x^2)*(9- x^2)^1/2dx. im having trouble ,any hint would be helpful.?

x = 3*sin(u), dx = 3*cos(u) du. Your integrand becomes

[9*sin^2(u)]*sqrt[9 - 9*sin^2(u)]*[3 cos(u)] du

= 81*sin^2(u)*cos(u)*cos(u) du

= (81/4)*sin^2(2u) du

= (81/8)*[1 - cos(4u)] du.

After integration you have

(81/8)*[u - (1/4)sin(4u)] + C.

To get back to "x" note that u = arcsin(x/3),

and sin(4u) = 2*sin(2u)*cos(2u)

= 4*sin(u)*cos(u)*[1 - 2*sin^2(u)]

= 4*(x/3)*sqrt[1 - (x/3)^2]*[1 - 2(x/3)^2].

"integral of integral of"

means there must be a second differential.

If the other differential was "y", just multiply "y" by your whole answer from above.

I really wanna know the point of asking here cause I don't get it?

How is that different from for example going to https://www.integral-calculator.com/ and taking hints (or the entire integral) from there, other than a difference in speed?

No double integrals with one dx :)