Find the indefinite integral: (6x3 - 3x + 1) dx

Use "k" for an arbitrary constant.
Dx and dx are not the same.
All the coefficients are integers.

separate into 3 integrals

integral(6x^3 dx)-integral(3x dx)+integral(1 dx)+k

6[integral(x^3 dx)]-3[integral(x dx)]+integral(1 dx)+k

6(x^4/4)-3(x^2/2)+x+k

(6x^4)/4-(3x^2)/2+x+k

Recall that the ∫x^(n) dx = x^(n + 1)/(n + 1) + k
In this case:

∫(6x^3 - 3x + 1) dx = 3x^4/2 - 3x²/2 + x + k where "k" is an arbitrary constant

The integral of 6x^3 - 3x + 1 dx
is (3/2)x^4 - (3/2)x^2 + x + k.