Evaluating lim x → 1 ∫ 1 x ln 2 ⁡ ( t ) d...

In terms of the evaluation of the limit: applying L'Hôpital's rule once forms
$\underset{x\to 1}{lim}\frac{{\int }_1^x{\ln }^2(t)dt}{(x-1{)}^3}=\underset{x\to 1}{lim}\frac{{\ln }^2(x)}{3(x-1{)}^2},$
where the chain rule implies that the derivative of the denominator is
$\frac{d}{dx}[(x-1{)}^3]=3(x-1{)}^2.$
From here you can apply L'Hôpital's rule two more times to find
$\underset{x\to 1}{lim}\frac{{\ln }^2(x)}{3(x-1{)}^2}=\underset{x\to 1}{lim}\frac{\ln (x)}{3x(x-1)}=\underset{x\to 1}{lim}\frac{1}{x(6x-3)}=\frac{1}{3}.$
To answer your second question, the result follows from the fundamental theorem of calculus. Let f be a continuous, real-valued function defined on the closed interval [a,b]. Suppose F is defined for all $x\in [a,b]$ by
$F(x)={\int }_a^{g(x)}f(t)dt.$
Then
$F^{\prime }(x)=\frac{d}{dx}{\int }_a^{g(x)}f(t)dt=f(g(x))g^{\prime }(x).$
Note that we could change the lower integration constant to any other positive real constant $a>0$ (as the fundamental theorem of calculus would still hold) to deduce
$F^{\prime }(x)=\frac{d}{dx}{\int }_a^x{\ln }^2(t)dt={\ln }^2(x).$
However, if the upper limit of integration was $x^2$ instead of x, then we would find
$F^{\prime }(x)=\frac{d}{dx}{\int }_a^{x^2}{\ln }^2(t)dt=2x{\ln }^2(x^2).$