How do I find the integral of x^2 (1 + 2x^3)^2 dx?

Answer 1

let, u = 1 + 2x^3
du = 6x^2 dx
du/6 = x^2 dx

substitute the above values in the integral,

integral{ (u)^2 du/6}
= 1/6 (u^3)/3 + c
= (u^3)/18 + c
= {1/18 (1 + 2x^3)^3} + c

there u go, use substitution.

Answer 2

Int x^2 (1 + 2x^3)^2 dx

Let u = 1 + 2x^3
Then du = 6x^2 dx

So, Int x^2 (1 + 2x^3)^2 dx
= Int 1/6 u^2 du
=1/6 (u^3)/3 +k
= (1 + 2x^3)/18 + k

Answer 3

I = ∫ x ² (1 + 2x ³) ² dx
let u = 1 + 2x ³
du = 6x ² dx
x ² dx = du / 6
I = (1/6) ∫ u ² du
I = (1/18) u ³ + C
I = (1/18) (1 + 2x ³) ³ + C

Answer 4

What. On. Earth.