Question

How do you integrate `int x^2 ln ^2x dx` using integration by parts?

Answer 1

`int x^2ln^2xdx = x^3/27(9ln^2x - 6lnx +2) +C`

Explanation:

We can integrate by parts using the logarithm as integral part, so that in the resulting integral we have a rational function:

`int x^2ln^2xdx = int ln^2x d(x^3/3)`

`int x^2ln^2xdx = (x^3ln^2x)/3 - 1/3 int x^3d(ln^2x)`

`int x^2ln^2xdx = (x^3ln^2x)/3 - 2/3 int x^3 lnx/xdx`

`int x^2ln^2xdx = (x^3ln^2x)/3 - 2/3 int x^2lnxdx`

Solve the resulting integral by parts again:

`int x^2lnxdx = int lnx d(x^3/3)`

`int x^2lnxdx = (x^3lnx)/3 - 1/3 int x^3 d(lnx)`

`int x^2lnxdx = (x^3lnx)/3 - 1/3 int x^3 dx/x`

`int x^2lnxdx = (x^3lnx)/3 - 1/3 int x^2 dx`

`int x^2lnxdx = (x^3lnx)/3 - 1/9x^3 +C`

Substituting in the first expression:

`int x^2ln^2xdx = (x^3ln^2x)/3 - 2/3( (x^3lnx)/3 - 1/9x^3 ) +C`

and simplifying:

`int x^2ln^2xdx = (x^3ln^2x)/3 - 2/9(x^3lnx) + 2/27x^3 +C`

`int x^2ln^2xdx = x^3/27(9ln^2x - 6lnx +2) +C`