Solve `int (ln^2(x))/x dx` ``

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`int (ln^2 x)/x dx`

We will assume that y= lnx:

`lnx = y ==> ln^2 x = y^2 `

`==> dy = 1/x dx ==> dx = xdy`

Now we will substitute into the integral:

`==> int (ln^2 x)/dx dx = int y^2/x xdy = int y^2 dy= y^3/3 + C`

But we know that y=...

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`int (log^2x)/x dx` =`int log^2 d log x= (log^3x)/x+C`

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