Integral calculus 2
posted by Tommy .
This is also one I can't figure out:
∫ x^3 (2x^2 + 1)^1/2 dx
u is obviously = 2x^2 + 1
du = 4x dx
x dx = 1/4 du
1/4 ∫ x^3 u^1/2
from here I know I can't integrate because I can't have more than 1 variable.
Thank you

I rewrote it as
∫ ( x(2x^2+1)^(1/2) (x^2) dx
I let u = x^2 and dv = x(2x^2 + 1)^(1/2) dx
du/dx = 2x
du = 2x dx
v = (1/6)(2x^2 + 1)^(3/2)
so we need: uv  ∫ v du
uv
= (1/6)(2x^2+1)^(3/2 (x^2)
= (x^2/6)(2x^2 + 1)^(3/2)
∫ (1/6)(2x^2 + 1)^(3/2) (2x) dx
= (1/30)(2x^2 + 1)^(5/2)
finally our integral
= (x^2/6)(2x^2 + 1)^(3/2)  (1/30)(2x^2 + 1)^(5/2)
= (1/30)(2x^2 + 1)^(3/2) [ 5x^2  (2x^2 + 1) ]
= (1/30)(2x^2 + 1)^(3/2) (3x^2  1)
wow, you better check my steps on this one. 
thank you