Does there exist a smooth projective surface $S / \mathbb{C}$ equipped with a dominant morphism $\pi: S \to \mathbb{P}^1$ which has the following properties:


1. The fibre at infinity is "multiple", i.e. as divisors on $S$ one has $\pi^{-1}(\infty) = mD$ for some $m > 1$ and some divisor $D$.
2. All other fibres are integral.



A pencil with exactly one multiple fibre
The answer is *yes*. 


In fact, it is possible to construct a rational elliptic fibration $f \colon S \to \mathbb{P}^1$ with exactly one multiple fibre of multiplicity $m \geq 2$, by starting from the blow-up of $\mathbb{P}^2$ at nine points that are the base locus of a pencil $\mathscr{P}$ of elliptic curves and then performing a logarithmic transformation centered at one point of $\mathbb{P}^1$.


Since the logarithmic transformation does not change the fibres outside the center, choosing a sufficiently general pencil $\mathscr{P}$ the reduced fibres of $f$ will be all irreducible.


For more details and examples, see 


Y. Fujimoto, [*On rational elliptic surfaces with multiple fibers*](http://dx.doi.org/10.2977/prims/1195171661), Publ. Res. Inst. Math. Sci. 26, No.1, 1-13 (1990). [ZBL0729.14027](https://zbmath.org/?q=an:0729.14027),


in particular Proposition 1.1.


