Hello, mathshelpneeded!

I can get you started on the second one . . .

∫ x²(1 + x²)^½ dx, .using the substitution: x = tanθ

We have: .x = tanθ . → . dx = sec²θ·dθ

. . and the radical becomes: secθ

Substitute: .∫ tan²θ·secθ·sec²θ·dθ . = . ∫ sec³θ·tan²θ·dθ

This integral is particularly annoying to integrate.

. . It has an odd-powered secant and an even-powered tangent.

It requires integration by parts and some fancy algebra.

I'll jump to the punchline:

. . (1/8)[2·sec³θ·tanθ + 3·secθ·tanθ + 3·ln|secθ + tanθ|] + C

and letyouback-substitute . . .