Use substitution and put `4x^3+2x = (x^4 + x^2)'`

Use integration by parts:

`int udv+ int vdu = uv`

Consider u = ln x=> du = `dx/x`

Consider `dv = 4x^3+2x = (x^4 + x^2)'=gt v = x^4 + x^2`

Use the formula of integration by parts:

`int (4x^3+2x)*ln x dx+ int ((x^4 + x^2)dx)/x = (x^4 + x^2)*ln x`

Subtract `int ((x^4 + x^2)dx)/x ` both sides:

`int (4x^3+2x)*ln x dx = (x^4 + x^2)*ln x - int ((x^4 + x^2)dx)/x`

You only need to calculate `int ((x^4 + x^2)dx)/x`

`int (x^4 + x^2)dx/x = int x^4 dx/x + int x^2dx/x`

`int (x^4 + x^2)dx/x = int x^3 dx + int x dx`

`int (x^4 + x^2)dx/x = x^4/4 + x^2/2` + c

**The result of integration is `int (4x^3+2x)*ln x dx = (x^4 + x^2)*ln x - x^4/4- x^2/2 + c` **

The intregral of (4x^3 + 2x)*ln x has to be found.

Let y = ln x

=> `x = e^y`

`x^3 = e^(3y)`

`dy/dx = 1/x = 1/e^y`

`int (4x^3 + 2x)*ln x dx`

=> `int (4*e^(3y) + 2*e^y)*y*e^y dy`

=> `int 4*e^(4y)*y + 2*y*e^(2y) dy`

=> `((4y - 1)*e^(4y))/4 +((2y - 1)*e^(2y))/2 `

Substitute y = ln x

=> `((4*lnx - 1)*x^4)/4 + ((2*ln x - 1)*x^2)/2 + C`

=> `(ln x(4*x^4 + 4x^2) - x^4 - 2*x^2)/4 + C`

**The required integral is : `(lnx(4x^4 + 4x^2) - x^4 - 2x^2)/4 + C` **