`int (ln(x))/x^4dx`

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

So let,

`u=ln(x) ` and `dv=1/x^4 dx = x^(-4 )dx`

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

Then, substitute u,v, and du to the formula.

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

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

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

`=- (x^(-3)ln(x))/3 - x^(-3)/9 + C`

`= -(ln(x))/(3x^3) - 1/(9x^3) + C`

**Hence, `int (ln(x))/x^4dx = - (ln(x))/(3x^3) - 1/(9x^3) + C` .**

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