`y=int((ln\ x)^2)/(x)\ dx`

Let `u= ln\ x`.

Then `du = 1/x dx`

`y=int((ln\ x)^2)/(x) dx= int u^2 du`

`=(u^3)/(3)+K`

`= ((ln\ x)^3)/(3)+K`

When `x=1`, `y=2`, so:

`2=((ln\ text[l])^3)/(3)+K`

`K=2`

Therefore the equation is

`y=((ln\ x)^3)/(3)+2`

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