Schwartz's QFT, in calculation of $(20.47)$

I am reading the Schwartz's Quantum field theory book, p.362~363, in particular derivation of $(20.47)$ and some question arises.

The following are associated preliminary formulas.

$\int d \Pi_{LIPS} = \frac{Q^2}{128 \pi^3} \int^{1-\beta}_{0} dx_1 \int^{1-\frac{\beta}{1-x_1}}_{1-x_1-\beta} dx_2 \tag{20.42}$

$\mathrm{Tr}[ \not ! {p_3}S^{\mu\alpha}\not!{p_4} S^{\alpha \mu}] = \frac{8e_R^2}{(1-x_1)(1-x_2) }\times {x_1^2 +x_2^2 + \beta[2(x_1+x_2) - \frac{(1-x_1)^2+(1-x_2)^2}{(1-x_1)(1-x_2)}] + 2 \beta^2 } \tag{20.43}$

$ \Gamma(\gamma^{\star} \to \mu^{+}\mu^{-}\gamma) = \frac{e_R^2}{2Q}\int d \Pi_{LIPS}\mathrm{Tr}[\not ! {p_3} S^{\mu \alpha} \not ! {p_4}S^{\alpha \mu}] \tag{1}$

And in p.363, he wrote, " The only terms that contribute as $\beta \to 0$ are

$ \int^{1-\beta}_{0} dx_1 \int^{1-\frac{\beta}{1-x_1}}_{1-x_1-\beta} dx_2 \frac{x_1^2+x_2^2}{(1-x_1)(1-x_2)} = \ln^2 \beta + 3 \ln \beta- \frac{\pi^2}{3}+6 + \mathcal{O}(\beta) \tag{20.45}$

and

$ -\beta \int^{1-\beta}_{0} dx_1 \int^{1-\frac{\beta}{1-x_1}}_{1-x_1-\beta} dx_2 \frac{(1-x_1)^2+(1-x_2)^2}{(1-x_1)^2(1-x_2)^2}= -1 + \mathcal{O}(\beta) \tag{20.46}$

Therefore, from $(20.42), (20.43), (1), (20.45), (20.46)$ ( $\beta:=\frac{m_{\gamma}^2}{Q^2}$ ),

$ \Gamma(\gamma^{\star} \to \mu^{+}\mu^{-}\gamma) = \frac{Qe_R^4}{32 \pi^3} {\ln^2 \frac{m_{\gamma}^2}{Q^2}+ 3 \ln \frac{m_{\gamma}^2}{Q^2}- \frac{\pi^2}{3} +5} \tag{20.47}$

Q. My question is, in derivation of $(20.47)$, why the term $\beta(2( x_1+x_2))$ in $(20.43)$ drops out as $\beta \to 0$?