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
To solve the integral ∫ x^3(2x^2 + 1)^(1/2) dx, you correctly identified the substitution u = 2x^2 + 1. Now, you differentiate both sides of this equation with respect to x:
du/dx = 4x
Next, you solve this equation for dx:
dx = du/(4x)
Substituting this value for dx into the original integral, we get:
∫ x^3(2x^2 + 1)^(1/2) dx = ∫ x^3(2x^2 + 1)^(1/2) (du/(4x))
Now, simplify the expression:
∫ (x^2(2x^2 + 1)^(1/2))/4 du
You now have an integral in terms of u only, so you can proceed to integrate with respect to u. The x variable has been eliminated.