Integration by parts and the logarithm function

The rule is correct. Here's some more information on integration by parts, including the proof. David's integral is correct, and here's how.

First integration by parts:

$\displaystyle u = \ln^2(x)$ , $\displaystyle dv = dx$ , $\displaystyle du = \frac{2\ln(x)}{x}$ , $\displaystyle v = x$

$\displaystyle \int \ln^2(x)$ = $\displaystyle x\ln^2(x) - 2\int \ln(x)$

Second integration by parts:

$\displaystyle u = \ln(x)$ , $\displaystyle dv = dx$ , $\displaystyle du = \frac{1}{x}$ , $\displaystyle v = x$

$\displaystyle 2\int \ln(x)$ = $\displaystyle x\ln(x)$ - $\displaystyle \int 1$

$\displaystyle 2\int \ln(x)$ = $\displaystyle 2x\ln(x)-2x$

$\displaystyle \int \ln^2(x)$ = $\displaystyle x\ln^2(x) - 2x\ln(x)+2x$

$\displaystyle \int \ln^2(x)$ = $\displaystyle x(\ln^2(x) - 2\ln(x)+2)$

Henry's answer is correct, but at first glance it might seem like we have a problem. $\displaystyle \ln(0)$ is undefined! If you graphed $\displaystyle \ln(x)$ , there'd be a vertical asymptote. And because $\displaystyle \ln(x)$ isn't continuous over the closed interval $\displaystyle [0,1]$ , $\displaystyle \ln^2(x)$ isn't either. There's still a solution, though. Let's split $\displaystyle x\ln^2(x) - 2\int \ln(x)$ into $\displaystyle x\ln^2(x)$ and $\displaystyle -2\int \ln(x)$ . Let's analyze $\displaystyle \lim_{x\to 0+} x\ln^2(x)$ . Note that this isn't 0, just very, very, close. We set up $\displaystyle x\ln^2(x)$ for L'Hopital's rule by swapping x for $\frac{1}{1/x}$ . Now we evaluate $\displaystyle \lim_{x\to 0+} \frac{\ln^2(x)}{1/x}$ . Let's use L'Hopital's rule, and take the derivative of the numerator and denominator. We end up with $\displaystyle \lim_{x\to 0+} \frac{1/x}{-1/x^2}$ , which we simplify to $\displaystyle \lim_{x\to 0+} -2x$ . We use direct substitution to get 0. Now we turn to what we are left with, or $\displaystyle -2\int \ln(x)$ . Let's use the expanded form of our integration by parts to get $\displaystyle (\lim_{x\to 0+} -2x\ln(x) )+2\int_{0}^{1} 1$ , and we use L'Hopital's rule the exact same way we did before, getting 0. We're left with $\displaystyle 2\int_{0}^{1} 1$ , which is simply $\displaystyle 2*1-2*0$ , or $\boxed{2}$ .