Genus of $y^4=x^{14}+x$? I want to calculate the genus of the algebraic curve $y^4=x^{14}+x$ by Riemann-Hurwitz formula.
I know there are $42$ branch numbers over all finite points by the discriminant of the curve. But i don't know how to calculate the branch number of infinity, because infinity is a singular point. I want to know how to resolve the singularities of infinity.
What's more, how can i use puiseux series to distinct the regular points, branch points and singular points? 
Someone help please? Thanks.

The curve is a smooth compactification the cyclic covering of $\mathbb{A}^1$ branched over the divisor
$$
D \subset \mathbb{A}^1
$$
defined by the equation $x^{14} + x = 0$. First, consider the double covering
$$
f \colon X \to \mathbb{P}^1
$$
branched over $D$. This is a hyperelliptic curve of genus $g$, where $2g + 2 = \deg(D) = 14$, i.e., $g = 6$. Note that $f^{-1}(D) = 2D'$, where $D' \subset X$ is a divisor of degree $14$. Note also that 
$$
\mathcal{O}_X(D') \cong f^*\mathcal{O}_{\mathbb{P}^1}(7)
$$
is not divisible by 2 in $\mathrm{Pic}(X)$. However, the divisor
$$
D'' = D' + f^{-1}(\infty)
$$
of degree 16 is divisoble by 2, because 
$$
\mathcal{O}_X(D'') \cong f^*\mathcal{O}_{\mathbb{P}^1}(8).
$$
Therefore, we can consider the double covering
$$
Y \to X
$$
branched over $D''$. Clearly, over $\mathbb{A}^1$ the curve $Y$ gives the required cyclic covering. Moreover, it is smooth. Hence its genus can be computed by the Hurwitz formula as 
$$
g(Y) = 2g(X) - 1 + \frac12 \deg(D'') = 2\cdot 6 - 1 + 8 = 19.
$$
