To evaluate the integral problem: `int_1^2 x^4ln(x) dx` , we follow the formula from basic integration table. For the integrals with logarithm, the problem resembles the formula:

`int x^n ln(x) dx = x^((n+1)) ( ln(x)/(n+1)- 1/(n+1)^2), n!= -1` .

By comparison of `x^n ` with `x^4` , we let `n=4`...

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To evaluate the integral problem: `int_1^2 x^4ln(x) dx` , we follow the formula from basic integration table. For the integrals with logarithm, the problem resembles the formula:

`int x^n ln(x) dx = x^((n+1)) ( ln(x)/(n+1)- 1/(n+1)^2), n!= -1` .

By comparison of `x^n ` with `x^4` , we let `n=4` which satisfy that condition n!=-1 to be able to use the aforementioned integral formula.

Then the integral problem is evaluated as:

`int_1^2 x^4ln(x) dx= [x^((4+1)) ( ln(x)/(4+1)- 1/(4+1)^2)]|_1^2`

` = [x^(5) ( ln(x)/5- 1/5^2)]|_1^2`

`= [x^(5) ( ln(x)/5- 1/25)]|_1^2`

`= [(x^(5) ln(x))/5- x^5/25]|_1^2`

Apply definite integral formula: `F(x)|_a^b = F(b) - F(a)` .

`[(x^(5) ln(x))/5- x^5/25]|_1^2=[(2^(5) ln(2))/5- 2^5/25]-[(1^(5) ln(1))/5- 1^5/25]`

`=[(32 ln(2))/5- 32/25]-[(1ln(1))/5- 1/25]`

`=(32 ln(2))/5- 32/25 -(1ln(1))/5+ 1/25`

`= (32 ln(2))/5 -0/5+ (1-32)/25`

`=( 32ln(2))/5 -31/25` or `3.196` (approximated value).