Integrate x(3x^2+2)^4 dx

To integrate x(3x^2+2)^4 dx, we can use the substitution method. Let u = 3x^2 + 2, then du = 6x dx. Rearranging, we get dx = du / 6x.

Now, substitute u = 3x^2 + 2 and dx = du / 6x into the integral:

∫ x(3x^2+2)^4 dx
= ∫ (u - 2)(u)^4 (du / 6x)
= ∫ (u^5 - 2u^4) du / 6
= (1/6) ∫ u^5 du - (1/3) ∫ u^4 du
= (1/6) * (u^6 / 6) - (1/3) * (u^5 / 5) + C
= u^6 / 36 - u^5 / 15 + C

Now, substitute back u = 3x^2 + 2:

= (3x^2 + 2)^6 / 36 - (3x^2 + 2)^5 / 15 + C

Therefore, the integral of x(3x^2+2)^4 dx is (3x^2 + 2)^6 / 36 - (3x^2 + 2)^5 / 15 + C.

Step 1: Expand the expression

(3x^2 + 2)^4 = (3x^2)^4 + 4(3x^2)^3(2) + 6(3x^2)^2(2)^2 + 4(3x^2)(2)^3 + (2)^4

= 81x^8 + 216x^6 + 216x^4 + 64x^2 + 16

Step 2: Integrate each term

∫x(81x^8 + 216x^6 + 216x^4 + 64x^2 + 16) dx

= ∫81x^9 dx + ∫216x^7 dx + ∫216x^5 dx + ∫64x^3 dx + ∫16x dx

= 9x^10 + 36x^8 + 36x^6 + 16x^4 + 8x^2 + C

Therefore, the integral of x(3x^2 + 2)^4 dx is 9x^10 + 36x^8 + 36x^6 + 16x^4 + 8x^2 + C.