Integrate pls.?

Question

⌡ x² √x³+ 2dx

Answer 1

⌡ x² √x³+ 2dx

= 1/3 ⌠ 3x² √x³ + 2dx

=1/3 ⌡√ x³ + 2(3x² dx)

=1/3(2/3) (x³+2)^3/2 + C

=2/9 (x³ +2)^3/2 + C

The 1(3/2) was written as 2/3, since this form is more convenient with fractions.

with u =x³+2 using u,

⌡ x² √x³+ 2dx

=⌡u^1/2(1/3du)

=1/3 ⌡u^ 1/2 du

=1/3(2/3) u^3/2 + C

=2/9u^3/2 +C

=2/9(x³+2)^3/2 + C

by substituting the value of u which is x³ + 2.

Answer 2

⌡ x² √(x³+ 2)dx

Let u = x^3 + 2, then du = 3x^2dx

So the desired integral is

⌡ (1/3)u^(1/2)du = (1/3)(2/3)u^(3/2) + C since 1/(3/2) = 2/3
= (1/2)u√u + C
So replacing back

⌡ x² √(x³+ 2)dx = (1/2)(x³+ 2)√(x³+ 2) + C

Answer 3

x² √x³+ 2 = x^(2 + 3/2) + 2 = x^7/2 + 2

Thus Integral = 2/9 x^9/2 + 2x + const

Answer 4

I am guessing that question should read:-
I = ∫ x² √(x³ + 2 ).dx which is of course different to above.
Let u = x³ + 2
du/dx = 3.x²
du/3 = x².dx
I = (1/3) ∫u^(1/2) du
I = (1/3).(2/3).u^(3/2) + C
I = (2/9).(x³ + 2)^(3/2) + C