Integrationn retarded question

integral x(3x^2-1)^4dx

∫x(3x² - 1)⁴ dx -----------> Let u = 3x² - 1

= ∫x(3x² - 1)⁴.(1/6x).du
= [1/6]∫u⁴ du
= [1/6](u⁵/5) + C
= [1/30](3x² - 1)⁵ + C

OR if you meant ∫x(3x¹)⁴ dx then,

∫x(3x¹)⁴ dx

= ∫ 81x⁵ dx
= 81∫x⁵ dx
= 81[x⁶/6] + C
= (81x⁶)/6 + C
= (27x⁶ / 2) + C

That's about it!

u substitution:

u=3x^2 -1
du=6x dx

the integral now is: integral (1/6)u^4 du = (1/30)u^5 + c = (1/30)(3x^2 -1)^5 + c

u = 3x^2 - 1
du = 6x * dx

(3x^2 - 1)^4 * x * dx =>
u^4 * (1/6) * du

Integrate

(1/6) * (1/5) * u^5 + C =>
(1/30) * (3x^2 - 1)^5 + C