# Solve `int (ln^2(x))/x dx` `` `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 +...

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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= ln x

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

``

Approved by eNotes Editorial Team Substitute ln x by other variable. Instead of ln x put the variable y, in this way: ln x=y => ln^2(x)=y^2

You will see that if differentiate ln x with respect to x the result is 1/x.

d(ln x)/dx=dy => (1/x)*dx=dy

You will integrate:

Integral (ln^2(x))/x dx=Integral y^2 dy=y^3/3 + constant

Substitute y by ln x in result.

Integral (ln^2(x))/x dx= (ln^3(x))/3 + constant

Answer: Integral (ln^2(x))/x dx= (ln^3(x))/3 + constant

Approved by eNotes Editorial Team