How do you do this??? I know how to do integration by parts it's just this one problem...
Evaluate the integral using integration by parts:
The integral of (lnx)^2 dx (that's lnx squared)
Thanks
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