$K(x)$ is a simple extension of $K(u)$ and find the dimension of it 
Let $K$ be a field, $x$ an indeterminate, and $u=\frac{x^4}{x−1} \in K(x)$.  Prove that $K(x)$ is a simple extension of $K(u)$.  What is $[K(x) :K(u)]$?

I want to find such an $\alpha \in K(x)$ such that $K(x)=K(u)(\alpha)$. I have a really problem with even starting the solution, I am thinking that this $\alpha$ should depend on $u$ but have no idea how to find it.
Edition:
Let $x$ be indeterminate. Then, we have that $p(t)=t^4-ut+u \in K(u)[t]$ and $p(x)=x^4-ux+u=0$, so $[K(x):K(u)]=4??$ I am not sure about that what is the way to show that $K(x)$ is simple in $K(u).$
 A: What you have done is that you established a surjective ring homomorphism $K(u)[t]/(p) \rightarrow K(x)$. This is a good starting point.
It remains to show that $p$ is irreducible in $K(u)[t]$, which would imply that the quotient $K(u)[t]/(p)$ is a field, hence the above ring homomorphism must be an isomorphism, and $K(x) \simeq K(u)[t]/(p)$ is by definition a simple extension (and obviously of degree $4$).
To show that $p$ is irreducible, there are many possible ways. The following is a more general result:

Proposition: Let $f, g \in K[t]$ be coprime polynomials. Then the polynomial $f + ug \in K(u)[t]$ is irreducible.

Proof: It is clear that $f + ug$, as a polynomial in $K[u][t]$, is irreducible, since it has degree $1$ in $u$, hence in any factorization $f + ug = pq$, one of $p$ and $q$ (say $p$) must have degree $0$ in $u$, i.e. is in fact a polynomial in $t$. It then follows that $p$ is a common factor of $f$ and $g$, which must be a constant.
The result then follows from Gauss's lemma for UFD.
