You need to integrate by parts, hence, you need to use the following formula such that:

`int f(x)g'(x)dx = f(x)g(x) - int f'(x)*g(x)dx`

Considering `f(x) = ln x`  and `g'(x) = x^3`  yields:

`f(x) = ln x => f'(x) = 1/x`

`g'(x) = x^3 => g(x) = int x^3 dx = x^4/4`

`int x^3*ln x dx = (x^4*ln x)/4 - int (1/x)(x^4/4)dx`

`int x^3*ln x dx = (x^4*ln x)/4 - (1/4)int x^3 dx ` `int x^3*ln x dx = (x^4*ln x)/4 - x^4/16 + c`

Factoring out `x^4/4`  yields:

`int x^3*ln x dx = (x^4/4)(ln x -1/4) + c`

Hence, evaluating the given indefinite integral using parts yields `int x^3*ln x dx = (x^4/4)(ln x - 1/4) + c` .

`int x^3 ln(x) dx`

To evaluate, use integration by parts. The formula is` int udv = uv - int vdu` . 

So let,

`u = ln (x) `                    and                  `dv= x^3 dx`

`du = 1/x dx `                                             `v=int x^3 dx=x^4/4`

Substitute u, v and du to the formula.

`int x^3 ln(x) dx = ln(x) * x^4/4 - int x^4/4 *1/xdx`

`int x^3 ln(x) dx=(x^4 ln(x))/4 - 1/4 int x^3 dx`

`int x^3 ln(x) dx=(x^4 ln(x))/4 - 1/4* x^4/4 + C`

`int x^3 ln(x) dx=(x^4 ln(x))/4 - x^4/16 + C`

Hence,  `int x^3 ln(x) dx=(x^4 ln(x))/4 - x^4/16 + C`  .