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

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Answer 1 · 2017-03-08T09:50:34

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

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:

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

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

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