You should evaluate the following limit to check if the integral diverges or converges such that:

`int_5^oo x*e^(-3x)dx`

You should use the following substitution such that:

`-3x = t => -3dx = dt => dx = -(dt)/3`

Changing the variable yields:

`int x*e^(-3x)dx = (1/9)int te^t dt`

You need to use integration by parts such that:

`int udv = uv - int vdu`

`u = t => du = dt`

`dv = e^t dt => v = e^t`

`int te^t dt = te^t - int e^t dt`

`int te^t dt = te^t - e^t + c`

`int te^t dt = e^t(t - 1) + c`

Substituting back -3x for t yields:

`int x*e^(-3x)dx = (1/9)e^(-3x)(-3x-1) + c`

You may evaluate the definite integral yields:

`int_5^oo x*e^(-3x)dx = lim_(n->oo) int_5^n x*e^(-3x)dx`

`int_5^oo x*e^(-3x)dx = lim_(n->oo) ((1/9)(e^(-3n)(-3n-1) - e^(-15)*(-16)))`

`lim_(n->oo) ((1/9)(e^(-3n)(-3n-1) - e^(-15)*(-16)))`

You should evaluate the limit `lim_(n->oo) (-3n-1)/(e^(3n)) ` such that:

`lim_(n->oo) (-3n-1)/(e^(3n)) = oo/oo`

Using l'Hospital's theorem yields:

`lim_(n->oo) ((-3n-1)')/((e^(3n))') = lim_(n->oo) -3/(3e^(3n)) = -3/oo = 0`

`lim_(n->oo) ((1/9)(e^(-3n)(-3n-1) - e^(-15)*(-16))) = 16e^(-15)/9`

Since the result of limit is finite, `lim_(n->oo) int_5^n x*e^(-3x)dx =16e^(-15)/9` , hence, the given improper integral converges.