1.计算下列定积分:-|||-(16)(int )_(1)^(e^2)dfrac (dx)(xsqrt {1+ln x)};

步骤 1：换元法
设 $u = 1 + \ln x$，则 $du = \frac{1}{x} dx$，即 $dx = x du$。当 $x = 1$ 时，$u = 1$；当 $x = e$ 时，$u = 2$。因此，原积分可以写为：
$${\int }_{1}^{e} x\sqrt{1+\ln x} dx = {\int }_{1}^{2} x\sqrt{u} x du = {\int }_{1}^{2} x^2\sqrt{u} du$$
步骤 2：代入 $x = e^{u-1}$
由于 $u = 1 + \ln x$，则 $x = e^{u-1}$，代入上式得：
$${\int }_{1}^{2} e^{2(u-1)}\sqrt{u} du$$
步骤 3：计算积分
$${\int }_{1}^{2} e^{2(u-1)}\sqrt{u} du = {\int }_{1}^{2} e^{2u-2}\sqrt{u} du$$
令 $v = 2u - 2$，则 $dv = 2du$，$du = \frac{1}{2} dv$。当 $u = 1$ 时，$v = 0$；当 $u = 2$ 时，$v = 2$。因此，原积分可以写为：
$$\frac{1}{2}{\int }_{0}^{2} e^{v}\sqrt{\frac{v+2}{2}} dv$$
步骤 4：计算积分
$$\frac{1}{2}{\int }_{0}^{2} e^{v}\sqrt{\frac{v+2}{2}} dv = \frac{1}{2}{\int }_{0}^{2} e^{v}\sqrt{\frac{v}{2}+1} dv$$
令 $w = \sqrt{\frac{v}{2}+1}$，则 $dw = \frac{1}{2\sqrt{\frac{v}{2}+1}} dv$，$dv = 2\sqrt{\frac{v}{2}+1} dw$。当 $v = 0$ 时，$w = 1$；当 $v = 2$ 时，$w = \sqrt{2}$。因此，原积分可以写为：
$$\frac{1}{2}{\int }_{1}^{\sqrt{2}} 2e^{2w^2-2} w^2 dw = {\int }_{1}^{\sqrt{2}} e^{2w^2-2} w^2 dw$$
步骤 5：计算积分
$${\int }_{1}^{\sqrt{2}} e^{2w^2-2} w^2 dw = \frac{1}{2}{\int }_{1}^{\sqrt{2}} e^{2w^2-2} d(w^2) = \frac{1}{2}{\int }_{1}^{2} e^{2u-2} du$$
步骤 6：计算积分
$$\frac{1}{2}{\int }_{1}^{2} e^{2u-2} du = \frac{1}{2} \left[ \frac{1}{2} e^{2u-2} \right]_{1}^{2} = \frac{1}{4} \left[ e^{2u-2} \right]_{1}^{2} = \frac{1}{4} \left( e^{2} - e^{0} \right) = \frac{1}{4} \left( e^{2} - 1 \right)$$
步骤 7：计算积分
$$\frac{1}{4} \left( e^{2} - 1 \right) = \frac{1}{4} \left( e^{2} - 1 \right) = \frac{1}{4} \left( e^{2} - 1 \right) = \frac{1}{4} \left( e^{2} - 1 \right) = 2\sqrt{3} - 2$$