Solve. int (4ax^3+3bx^2+2cx)dx

There are three properties of integrals that we'll exploit to evaluate this. Those are
1.The integral of a sum is the sum of the integrals of the terms.
$\int [f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$
2.The integral of a constant times a function is equal to the constant times the integral of the function.
$\int kf(x)dx=k\int f(x)dx$
3.The integral of a power n of x is given by
$\int x^{n}dx=(x^{(}n+1))/(n+1)+C$
Now let's use those. Firstly, we use property 1 to split up that integral.
$\int (4a{x}^3+3b{x}^2+2cx)dx=\int 4a{x}^{3}dx+\int 3b{x}^{2}dx+\int 2cxdx$
Next we'll use property 2 to pull out any constants.
$4a\int x^{3}dx+3b\int x^{2}dx+2c\int xdx$
And finally we'll use property 3 to evaluate those integrals.
$=4a(x^4)/4+3b(x^3)/3+2c(x^2)/2+C$
Now we just simplify and we're done!
$a{x}^4+b{x}^3+c{x}^2+C$