= Solution
For fixed $0<\gamma<1$, the denominator's first squared term has no zero on the interval and is at least $(1-\gamma)^2$. Expand the integrand uniformly in $\varepsilon^2$ and integrate:
$$
I(\gamma;\varepsilon)=\int_0^1\frac{dx}{[1-\gamma+\gamma x]^2}+O(\varepsilon^2)
=\boxed{\frac1{1-\gamma}+O(\varepsilon^2).}
$$
The <asymptotic expansion> is uniform for $\gamma$ in compact subsets below $1$, including the continuous limit $\gamma\to0$. It is not uniform through $\gamma=1$, where a narrow endpoint peak forms. That transition occurs when $1-\gamma=O(\varepsilon)$, as the next part shows.
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