i've been working on a formula for numerical integration envolving gaussian quadracture but i need to calculate the weights and abcisses in a given formula for any given N.
I read that the weights follow this formula but i dont really understand what P'n(xi)² is.
There are a variety of "Gaussian quadrature" methods varying in the interval of integration and an optional weighting function.
The weight you cite is for Gauss-Legendre quadrature. Here, $P_n'(x_i)$ is the first derivative of the $n^\text{th}$ Legendre polynomial. There are various formulae for expressing the Legendre polynomials directly. Differentiating a polynomial is straightforward.
The $x_i$ are the zeroes of the $P_n(x)$ polynomials. The $P_n(x)$ polynomial has $n$ zeroes and they exhibit reflection symmetry through the point $ x= 0$. (Consequently, $x = 0$ is always a root of $P_n$ when $n$ is odd.) The interlacing property says that the $n$ roots of $P_n(x)$ partition $(-1,1)$ into intervals, each of which contains exactly one zero of $P_{n+1}(x)$. This can be used to numerically find the zeroes of successive $P_n(x)$s. I mention numerically finding the zeroes because algebraically solving for the zeroes of polynomials of high degree is infeasible.
So say we are working with $n = 5$ (that is, our integration formula will evaluate the integrand $5$ times). The $x_i$ are the zeroes of $P_5(x)$, which are the zeroes of $\frac{1}{8}(63x^5 - 70 x^3 + 15x)$. Let's number the $x_i$ in order from least to greatest, so $x_1 < x_2 < \cdots < x_5$. We compute $P'_5(x) = \frac{1}{8}(315x^4 - 210x^2 + 15)$. Those zeroes are $x_1 = -\frac{1}{3}\sqrt{5+ 2\sqrt{\frac{10}{7}}}$, $x_2 = -\frac{1}{3}\sqrt{5- 2\sqrt{\frac{10}{7}}}$, $x_3 = 0$, $x_4 = \frac{1}{3}\sqrt{5- 2\sqrt{\frac{10}{7}}}$, and $x_5 = \frac{1}{3}\sqrt{5+ 2\sqrt{\frac{10}{7}}}$. (That is, $-0.90618{\dots}$, $-0.538469{\dots}$, $0$, $0.538469{\dots}$, and $0.90618{\dots}$.)
Then \begin{align*} w_1 = w_5 &= \frac{2}{(1-x_1^2)[P'_5(x_1)]^2} \\ &= \frac{2}{(1-(-0.90618{\dots})^2)[\frac{1}{8}(315(-0.90618{\dots}))^4 - 210(-0.90618{\dots}))^2 + 15)]^2} \\ &= \frac{322-13\sqrt{70}}{900} \text{,} \end{align*} assuming no typographical or computational errors. The weight $w_2 = w_4 = \frac{322+13\sqrt{70}}{900}$. The weight $w_3$ (for $x_3 = 0$) is $128/225$.
Generally, the $x_i$ are sparse in the middle of the interval and get denser near the endpoints. The weights tend to be higher in the middle and lower near the ends. (This pattern of weights makes sense: samples near the middle represent wider intervals and samples near the ends represent narrower intervals. If the integrand were constant, the weights would have to be greater in the middle and less near the ends to capture the widths of the area rectangles represented by the samples.)
(In response to a comment by OP.)
One can get the Legendre polynomials using Rodriguez's formula, $$ P_n(x) = \frac{1}{2^n n!} \frac{\mathrm{d}^n}{\mathrm{d}x^n}\left[ (x^2-1)^n \right] \text{,} $$ but unless you have a symbolic differentiator, this is not a great plan. A much better plan is to use any of the other formulas at the link above. For instance, from $$ P_n(x) = \frac{1}{2^n}\sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n}{k} \binom{2n-2k}{n} x^{n-2k} $$ we obtain $$ P_n'(x) = \frac{1}{2^n}\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} (-1)^k (n-2k)\binom{n}{k} \binom{2n-2k}{n} x^{n-2k-1} \text{.} $$ Where we have used standard notation for the floor function.