Clever way of evaluating $\frac{1}{\sqrt{1-2xt+x^2}} dt$ How would you spot a good substitution here?
Or perhaps there is a better way of solving it (i.e. without a substitution).
It is misleadingly (unless you can in fact arrive at useful result) attractive to factorize it: $$\int_{-1}^1\frac{1}{\sqrt{(x-t)^2-(t^2-1)}}dt$$ but the factorization does not seem to lead anywhere.
I shall not say what the substitution is that is used in the result in hope that there is a clever way of arriving at it that you will inform me of.

Hint: When you differentiate $\sqrt{t}$ you obtain a root in the denominator. So, try to do the same with the term in the square root (differentiate with respect to $t$)
$$\left(\sqrt{1-2xt+x^2}\right)'=\frac{1}{2\sqrt{1-2xt+x^2}}(1-2xt+x^2)'=\frac{-x}{\sqrt{1-2xt+x^2}}$$ and $x$ is a constant.
