simplify and state restriction
$$\frac{x}{x^2-3x-4}-\frac{4}{x+1}$$

Step 1 ：Factor the quadratic in the denominator of the first fraction. The quadratic $x^2-3x-4$ can be factored into $(x-4)(x+1)$. So, the expression becomes: $$\frac{x}{(x-4)(x+1)}-\frac{4}{x+1}$$

Step 2 ：Find a common denominator for the two fractions. In this case, the common denominator is $(x-4)(x+1)$. So, we rewrite the second fraction with this common denominator: $$\frac{x}{(x-4)(x+1)}-\frac{4(x-4)}{(x-4)(x+1)}$$

Step 3 ：Combine the numerators since the fractions have the same denominator: $$\frac{x-4x+16}{(x-4)(x+1)}$$

Step 4 ：Simplify the numerator to get: $$\frac{-3x+16}{(x-4)(x+1)}$$

Step 5 ：The restrictions are the values of $x$ that make the denominator equal to zero, because division by zero is undefined. Setting each factor in the denominator equal to zero gives: $x-4=0$ --> $x=4$ and $x+1=0$ --> $x=-1$. So, the restrictions are $x\ne 4$ and $x\ne -1$.

Step 6 ：So, the simplified form of the given expression is $$\boxed{{\displaystyle \frac{-3x+16}{(x-4)(x+1)}}}$$ with restrictions $x\ne 4$ and $x\ne -1$.