Let $X_1,X_2,...,X_n \sim Unif[0,\beta]$ and $\hat{\beta_1} = (n+1)X_{min}$ and $\hat{\beta_2} = (n+1)/n X_{max}$. Compute $\frac{var[\hat{\beta_1}]}{var[\hat{\beta_2}]}$. What I have done is the following:
$$ \begin{align*} \frac{var[\hat{\beta_1}]}{var[\hat{\beta_2}]} & = \frac{var[(n+1)X_{min}]}{var[\frac{n+1}{n}X_{max}]} \\ & = \frac{var[(n+1) n \left(1 - \frac{x}{\beta} \right)^{n-1} \frac{1}{\beta}]}{var[\frac{n+1}{n} \frac{n}{\beta^n} x^{n-1}]} \\ & = \frac{\int _0^{\beta \:}\:x^2\:\left(n+1\right)\:n\:\left(1\:-\:\frac{x}{\beta \:}\:\right)^{n-1}\:\frac{1}{\beta }dx-\beta ^2}{\int _0^{\beta \:}\:x^2\:\frac{n+1}{n}\frac{n}{\beta ^n}x^{n-1}dx-\beta ^2} \\ & = \frac{\frac{\beta^2n^2 + \beta^2n}{n+2} - \beta^2n}{\frac{\beta^2(n+1)}{n+2}-\beta^2} \\ & = n \end{align*} $$
However, I am not sure my final answer is correct and if it is what do I conclude? That the ratio variance between two estimators is equal to the amount of samples?
Your final answer is not quite correct (somehow both the variances you found look negative):
Both estimators are unbiased
$X_{\min}$ and $X_{\max}$ have the same variance of $\frac{n}{(n+1)^2(n+2)}\beta^2$, which can be found from a Beta distribution
so $(n+1)X_{\min}$ has a variance of $\frac{n}{(n+2)}\beta^2$
and $\frac{n+1}{n}X_{\max}$ has a variance of $\frac{1}{n(n+2)}\beta^2$
giving a ratio of the variances of $n^2$
So the ratio of variances between the two estimators is equal to the square of the sample size. Using the estimator based on the sample maximum looks much better, which makes sense as the maximum is the minimal sufficient statistic here.