Question

Find `int ln^2x dx`?

Answer 1

`int \ ln^2x \ dx = x^2lnx-2xlnx+2x + C`

Explanation:

We seek:

`I = int \ ln^2x \ dx`

We can apply integration by Parts:

Let `{ (u,=ln^2x, => (du)/dx,=(2lnx)/x), ((dv)/dx,=1, => v,=x ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

We have:

`int \ (ln^2x)(1) \ dx = (ln^2x)(x) - int \ (x)((2lnx)/x) \ dx`
`:. I = x^2lnx-2int \ lnx \ dx`

Now consider the last integral:

`I_1 = int \ x^2sin2x \ dx`

We can again apply integration by parts a second time for the second integral:

Let `{ (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=1, => v,=x ) :}`

Then plugging into the IBP formula:

`int \ (lnx)(1) \ dx = (lnx)(x) - int \ (x)(1/x) \ dx`
`:. int \ lnx \ dx = xlnx - int \ dx`
`:. int \ lnx \ dx = xlnx - x`

Using this result along with the earlier result we find:

`I = x^2lnx-2(xlnx-x) + C`
`\ \ = x^2lnx-2xlnx+2x + C`