let $k[x]$ be the ring of polynomial over a field $k$. Let $f \in k[x]$ an irreducible polynomial. Then $k[x][\frac{1}{f}]$ is not integral over $ k[x]$.
I solved this ex. In this way By absurd, $\frac{1}{f}$ is integral over $k[x]$, so we have a relation $$ \frac{1}{f^n} = a_{n-1}\frac{1}{f^{n-1}} + \cdots + a_1 \frac{1}{f} + a_0 $$ with $a_i \in k[x] \ \forall i \in \lbrace 0 , \dots , n-1 \rbrace $.
the right part of the equation can be written as $$ \frac{ a_{n-1} + a_{n-2}f + \cdots a_0 f^{n-1}}{f^{n-1}}$$ which is equal to $$ \frac{b}{f^{n-1}}$$ So we obtained $$ \frac{1}{f^n} = \frac{b}{f^{n-1}}$$ which is equivalent to $$\frac{1}{f} = b \ \text{ or } \ 1=fb$$ but this is impossible because $\text{deg}(f) \geq 1$
I didn't use he irreducibility of the polynomial $f$, so maybe I've done a mistake. Can someone explain me if the hypothesis is necessary?
An irreducible polynomial in $k[x]$ is usually defined as one which is not a unit and cannot be written as the product of two non-unit polynomials in $k[x]$. This is included in the hypothesis of your exercise to exclude polynomials of degree $0$. As you have shown, $$f^{-n} = a_{n-1}f^{-(n-1)}+\cdots+a_0 \implies\\ 1=fb$$ Which implies that $f$ is not irreducible.