F(x) = antideriv x(x^2 + 2)^5 dx?

Question

Give a specific function for F(x).
F(x) = ______________ + C.

Answer 1

∫ x(x^2 + 2)^5 dx=...(*)

x^2+2=u
2xdx=du
xdx=1/2du =>
(*)=1/2 ∫u^5du=
(1/2)(u^6/6)+C=
(x^2+2)^6/12 + C

Answer 2

make a substitution, u = x^2 + 2.

du = 2x dx

Therefore, x dx = du/2:

antiderivative (1/2) * (u^5) du. This is easy to integrate:

(1/2) * (1/6) * (u^6) = 1/12 * (u^6). Plug u back in:

F(x) = (1/12) * (x^2 + 5)^6 + C

Answer 3

I = ∫ x (x² + 2)^5 dx
Let u = x² + 2
du = 2x dx
du / 2 = x dx
I = (1/2) ∫ u^5 du
I = (1/2) u^6 / 6 + C
I = (1/12) (x² + 2)^6 + C

Answer 4

let, x^2 = u
du = 2x dx
du/2 = dx

antiderivative( (u + 5)^5 du/2 )

= 1/2*[(u + 5)^6]/6
= ((u+5)^6)/12 + c
= ((x^2 + 5)^6)/12 + c

Answer 5

Pie Squared ! 3.18278548\2