It can be decomposed into the form
$$\frac{x^2}{(x+\alpha)(x-\beta)^2} = \frac{\lambda}{x+\alpha} + \frac{\mu}{x-\beta} + \frac{\nu}{(x-\beta)^2}$$
which means
$$ \lambda(x-\beta)^2 + \mu(x+\alpha)(x-\beta) + \nu(x+\alpha) = x^2
$$
It can be written as a set of equations
$$\begin{align*}
\lambda + \mu &= 1 \\
-2\beta\lambda + (\alpha - \beta)\mu + \nu &= 0 \\
\beta^2\lambda - \alpha\beta\mu + \alpha\nu &= 0
\end{align*}$$
Solving for $\lambda, \mu, \nu$ will give you the correct decomposition. (I used CAS here)
$$
\left[\begin{array}{ccc|c}
1 & 1 & 0 & 1 \\
-2\beta & \alpha-\beta & 1 & 0 \\
\beta^2 & -\alpha\beta & \alpha & 0
\end{array}\right]
\rightarrow
\left[\begin{array}{ccc|c}
1 & 0 & 0 & \frac{\alpha^2}{(\alpha+\beta)^2} \\
0 & 1 & 0 & -\frac{\beta^2+2\alpha\beta}{(\alpha+\beta)^2} \\
0 & 0 & 1 & \frac{\beta^2}{\alpha+\beta}
\end{array}\right]
$$
Hence
$$ \lambda = \frac{\alpha^2}{(\alpha+\beta)^2}, \mu = -\frac{\beta^2+2\alpha\beta}{(\alpha+\beta)^2}, \nu = \frac{\beta^2}{\alpha+\beta}$$