Find # int ln^2x dx #?

1 Answer
Nov 28, 2017

# 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 #