Hello, abcocoa!

Did you try it at all? .There aren't many choices.

∫ (ln x)² dx

. . . . . . . . . . . . . . . . . . . . 2 ln x

Let u .= .(ln x)² . → . du .= .------- dx

. . . . . . . . . . . . . . . . . . . . . .x

dv .= .dx . → . v .= .x

And we have: . x(ln x)² - 2 ∫ ln x dx

. . By parts again:

. . . . u .= .ln x . → . du .= .dx/x

. . . . dv .= .dx/x . → . v .= .x

And we have: . x(ln x)² - 2(x·ln x - ∫ dx) + C

. . . . . . . . . . x(ln x)² - 2x·ln x + 2x + C