Limit and Continuity Find the limit (if it exists) and discuss the continuity of

What is the proof of the following:
${\displaystyle {\int }_0^1{\left(\frac{\ln t}{1-t}\right)}^{2}dt=\frac{{\pi }^2}{3}}$ ?

How do you find the coordinates of the points on the curve ${\displaystyle x^2-xy+y^2=9}$ where the tangent line is vertical?

Evaluate the folowing limits.
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2-7x+1}-x)}$
${\displaystyle \underset{x\to -\mathrm{\infty }}{lim}(\sqrt{x^2-7x+1}-x)}$

I know that the above limit is 1 from Wolfram but I had no proper way to prove. I attempted a proof and it is as follows:
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left({\displaystyle \frac{1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}+\cdots +{\displaystyle \frac{1}{n}}}{\log n}}\right)={\displaystyle \frac{{\displaystyle {\int }_0^1{\displaystyle \frac{1}{n}}dn}}{{\displaystyle \underset{n\to \mathrm{\infty }}{lim}\log n}}}.}$

Prove that $\underset{\alpha \to 0^{+}}{lim}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{n^{\alpha }}=1/2$

Uniform continuity and antiderivative of $x\mapsto e^{-x^2}$
Show that the antiderivatives of $x\mapsto e^{-x^2}$ are uniformly continuous in $\mathbb{R}$.
So we know that for a function to be uniformly continuous there has to exists $\xi$ s.t when $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$.
But how do we go about this since we cannot really integrate $e^{-x^2}$ and we need the antiderivative?