integrate 3x√1-2x^2 dx?

Question

integrate 3x√1-2x^2 dx
please show all work thank you!!!

show all work please! 1-2x^2 is all under the square root thank you!

Answer 1

I = ∫ 3x (1 - 2x ²)^(1/2) dx
let u = 1 - 2x ²
du = - 4x dx
- du / 4 = x dx
- (3/4) du = 3x dx
I = (- 3/4) ∫ u ^(1/2) du
I = (- 3/4) u^(3/2) / (3/2) + C
I = (- 1 / 2) (1 - 2x²)^(3/2) + C

Answer 2

whoops! took the derivative the first time I answered.

let's define a function u(x) to help us
u = 1-2x^2
du = -4x dx

so integral of 3x*sqrt(1-2x^2)dx can be written in terms of u as

integral of -3/4 sqrt(u) du
let { stand for taking the integral since I can't make the real sign for it

{ (-3/4) sqrt(u) du
(-3/4) { sqrt(u) du
(-1/2) u^(3/2) + c

plugging back in for u = 1-2x^2
(-1/2)(1-2x^2)^(3/2) + c

let's check by taking the derivative
(-1/2)(1-2x^2)^(3/2) + c
(1/2)(-3/2)(1-2x^2)^(1/2)(-4x)
3x * sqrt(1-2x^2)
yep!

Answer 3

Integral (3x*sqrt(1-2x^2)) dx

Note that
3x dx = (3/2) * d(x^2)
= (3/4) * d(2x^2)
= (3/4) * (- d((1 - 2x^2)))

So let u(x) = 1 - 2x^2

Then
3x dx = -(3/4) du
sqrt(1 - 2x^2) = sqrt(u)

and the integral is equal to:
Integral (-(3/4)sqrt(u)) du
= - (3/4) Integral (sqrt(u)) du
= - (3/4) Integral u^(0.5) du
= - (3/4) (u^(1.5))/(1.5)
= - (3/(4*1.5)) u^(1.5)
= - (1/2) u^(1.5)
= - (1/2) (1 - 2x^2)^(1.5)

So the answer is:
the integral = - (1/2) (1 - 2x^2)^(1.5)
= - (1/2) (sqrt(1 - 2x^2))^3