You need to integrate the functions `y = xe^(4x)` and `y = xe^(5x)` using parts such that:

`int xe^(4x) dx = x*(e^(4x))/4 - int (e^(4x))/4 dx `

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

`g'(x) = e^(4x) => g(x) = (e^(4x))/4`

`int xe^(4x) dx = x*(e^(4x))/4 - (e^(4x))/16 + c`

Reasoning by analogy yields:

`int xe^(5x) dx = x*(e^(5x))/5 - (e^(5x))/25 dx + c`

**Hence, evaluating the antiderivatives of the given functions yields `int xe^(4x) dx = x*(e^(4x))/4 - (e^(4x))/16 + c` and `int xe^(5x) dx = x*(e^(5x))/5 - (e^(5x))/25 dx + c` .**

To get the antiderivative of y = x*e^4x, we'll integrate by parts, so, we'll recall the formula:

Int udv = u*v - Int vdu (*)

We'll put u = x. (1)

We'll differentiate both sides:

du = dx (2)

We'll put dv = e^4x (3)

We'll integrate both sides:

Int dv = Int e^4x dx

v = e^4x/4 (4)

We'll substitute (1) , (2) , (3) and (4) in (*):

Int udv = x*e^4x/4 - Int (e^4x/4)dx

The anti-derivative of y = x*e^4x is:

**Int (x*e^4x)dx = (x*e^4x)/4 - (e^4x)/16 + C**

We'll calculate in the same way the antiderivative of y = x*e^5x

We'll put u = x. (5)

We'll differentiate both sides:

du = dx (6)

We'll put dv = e^5x (7)

We'll integrate both sides:

Int dv = Int e^5x dx

v = e^5x/5 (8)

We'll substitute (5) , (6) , (7) and (8) in (*):

Int udv = x*e^5x/5 - Int (e^5x/5)dx

**Int (x*e^4x)dx = (x*e^5x)/5 - (e^5x)/25 + C**