Double integral of projectile problem

Given the projectile problem

$$x^{\prime \prime}=-\frac{1}{(1+\epsilon x)^{2}}$$

$$x(0) = 0 \quad x^{\prime}(0) = 1$$

By integrating twice, the projectile problem can be rewritten as the integral equation

$$x(t) = t - \int^{t}_{0}\int^{\tau}_{0}\frac{ds}{(1+\epsilon x(s))^{2}}d\tau = t - \int^{t}_{0} \frac{(t- \tau)d\tau}{(1+\epsilon x(\tau))^{2}}$$

The previous was a statement I found in my applied analysis notes. I understand what we are doing, but I'm not sure how the professor managed to reach the second result of the double integral.