Let $\alpha$ and $\beta$ be the real numbers such that $\lim_{x \to 0} \frac{1}{x^{3}} \left( \frac{\alpha}{2} \int_{0}^{x} \frac{1}{1-t^{2}}\,dt + \beta x \cos x \right) = 2.$ Then the value of $\alpha + \beta$ is ________.

$\lim_{x\to 0}\dfrac{1}{x^{3}}\left(\dfrac{\alpha}{2}\int_{0}^{x}\dfrac{dt}{1-t^{2}}+\beta x\cos x\right)=2$ $\text{Find } \alpha+\beta$ $1.\ \text{expand both terms in powers of } x$ $2.\ \dfrac{N(x)}{x^{3}} \to \text{finite}$ $3.\ \text{solve for } \alpha,\ \beta$ $\dfrac{1}{1-t^{2}}=1+t^{2}+t^{4}+\cdots$ $\int_{0}^{x}\dfrac{dt}{1-t^{2}}=x+\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}+\cdots$ $x\cos x=x\left(1-\dfrac{x^{2}}{2}+\dfrac{x^{4}}{24}-\cdots\right)=x-\dfrac{x^{3}}{2}+\cdots$ $N(x)=\dfrac{\alpha}{2}\left(x+\dfrac{x^{3}}{3}+\cdots\right)+\beta\left(x-\dfrac{x^{3}}{2}+\cdots\right)$ $=\left(\dfrac{\alpha}{2}+\beta\right)x+\left(\dfrac{\alpha}{6}-\dfrac{\beta}{2}\right)x^{3}+O(x^{5})$ $\dfrac{N(x)}{x^{3}}=\dfrac{\alpha/2+\beta}{x^{2}}+\left(\dfrac{\alpha}{6}-\dfrac{\beta}{2}\right)+O(x^{2})$ $\text{limit finite} \Rightarrow \dfrac{\alpha}{2}+\beta=0 \Rightarrow \beta=-\dfrac{\alpha}{2}$ $\dfrac{\alpha}{6}-\dfrac{\beta}{2}=2$ $\dfrac{\alpha}{6}+\dfrac{\alpha}{4}=\dfrac{5\alpha}{12}=2 \Rightarrow \alpha=\dfrac{24}{5}$ $\beta=-\dfrac{12}{5}$ $\alpha+\beta=\dfrac{24}{5}-\dfrac{12}{5}=\dfrac{12}{5}$ $\boxed{\alpha+\beta=\dfrac{12}{5}}$