.计算下列定积分:-|||-(10) (int )_(1)^sqrt (3)dfrac (dx)({x)^2sqrt (1+{x)^2}}

步骤 1：换元法
令 $x = \tan\theta$，则 $dx = \sec^2\theta d\theta$，且当 $x = 1$ 时，$\theta = \frac{\pi}{4}$；当 $x = \sqrt{3}$ 时，$\theta = \frac{\pi}{3}$。因此，原积分变为
$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sec^2\theta d\theta}{\tan^2\theta \sqrt{1 + \tan^2\theta}} = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sec^2\theta d\theta}{\tan^2\theta \sec\theta} = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{d\theta}{\sin^2\theta} = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \csc^2\theta d\theta
$$
步骤 2：积分
根据积分公式 $\int \csc^2\theta d\theta = -\cot\theta + C$，我们有
$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \csc^2\theta d\theta = -\cot\theta \Big|_{\frac{\pi}{4}}^{\frac{\pi}{3}} = -\cot\frac{\pi}{3} + \cot\frac{\pi}{4} = -\frac{1}{\sqrt{3}} + 1 = 1 - \frac{\sqrt{3}}{3}
$$
步骤 3：化简
将上一步的结果化简，得到
$$
1 - \frac{\sqrt{3}}{3} = \frac{3}{3} - \frac{\sqrt{3}}{3} = \frac{3 - \sqrt{3}}{3} = \frac{2\sqrt{3}}{3} - \sqrt{2}
$$