I have a problem with the computation of Lobatto quadrature formula. In the second snippet of the Czech textbook they say:
we directly compute $(9.23)$ but I do not follow how did they computed $(9.25)$ and $$A_i=\frac{\bar{A}_i}{1-x_i}^2, i=1,...,n-1$$
I think that integral of $P_1(x)$ is $$-\frac{(1-x)^2}{4}f(-1)+\frac{(1+x)^2}{4}f(1)$$ and hence form $-1$ to $1$ is $f(-1)+f(1)$; what next, how do I compute $A_i$ in $(9.25)$ ?
In 9.23 you insert some other quadrature method for the error of the trapezoidal method $$ \int_{-1}^1(f(x)-P_1(x))\,dx=\int_{-1}^1 g(x)w(x)\,dx\approx\sum_{k=1}^{n-1}\bar A_kg(x_k),\\~~~where~~~g(x)=\frac{f(x)-P_1(x)}{1-x^2}, $$ so that in total you get $$ I(f)\approx f(-1)+f(1)+\sum_{k=1}^{n-1}\bar A_kg(x_k) =f(-1)+\sum_{k=1}^{n-1}\bar A_k\frac{f(x_k)-P_1(x_k)}{1-x_k^2}+f(1), $$ then separate the fraction with $$ \frac{P_1(x_k)}{1-x_k^2}=\frac1{2(1+x_k)}f(-1)+\frac1{2(1-x_k)}f(1) $$ and compare coefficients of $f(x_k)$, $k=0,...,n$ of the resulting expression with 9.25. Note that 9.25 is just a restatement of the target formula for $I(f)\approx Q(f)$.