One. (20 points) Given the equation in $x$
$$
(m+1) x^{2}+2 x-5 m-13=0
$$

has roots that are integers. Find the value of the real number $m$.
When $m=-1$, the root of the equation $x=4$ satisfies the condition.

When $m \neq-1$, let the two roots of the equation be $x_{1}$ and $x_{2}$, and $x_{1}>x_{2}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{2}{m+1}, \\
x_{1} x_{2}=\frac{-5 m-13}{m+1}
\end{array}\right. \\
\Rightarrow\left(x_{1}-4\right)\left(x_{2}-4\right)=11=11 \times 1=(-11) \times(-1) \\
\Rightarrow\left(x_{1}-4, x_{2}-4\right)=(11,1) \text { or }(-1,-11) \\
\Rightarrow\left(x_{1}, x_{2}\right)=(15,5) \text { or }(3,-7) \\
\Rightarrow m=-\frac{11}{10} \text { or }-\frac{1}{2} .
\end{array}
$$

In summary, the real number $m=-1,-\frac{11}{10}$ or $-\frac{1}{2}$.