(int )_(0)^1dfrac (x+2)(sqrt {x+1)}dx=( )(int )_(0)^1dfrac (x+2)(sqrt {x+1)}dx=( );(int )_(0)^1dfrac (x+2)(sqrt {x+1)}dx=( );(int )_(0)^1dfrac (x+2)(sqrt {x+1)}dx=( );(int )_(0)^1dfrac (x+2)(sqrt {x+1)}dx=( );

步骤 1：拆分被积函数
将被积函数拆分为两个部分，以便于分别计算积分。
${\int }_{0}^{1}\dfrac {x+2}{\sqrt {x+1}}dx={\int }_{0}^{1}\dfrac {x}{\sqrt {x+1}}dx+{\int }_{0}^{1}\dfrac {2}{\sqrt {x+1}}dx$

步骤 2：变量替换
对于${\int }_{0}^{1}\dfrac {x}{\sqrt {x+1}}dx$，令$\sqrt {x+1}=t$，则$x={t}^{2}-1$，$dx=2tdt$，代入积分式中。
${\int }_{0}^{1}\dfrac {x}{\sqrt {x+1}}dx=2{\int }_{1}^{\sqrt {2}}\dfrac {{t}^{2}-1}{t}\cdot tdt=2{\int }_{1}^{\sqrt {2}}({t}^{2}-1)dt$

步骤 3：计算积分
计算上述积分，得到结果。
$2{\int }_{1}^{\sqrt {2}}({t}^{2}-1)dt=2(\dfrac {1}{3}{t}^{3}{|}_{1}^{\sqrt {2}}-t|{1}^{\sqrt {2}})=2(\dfrac {2}{3}\sqrt {2}-\dfrac {1}{3}-\sqrt {2}+1)$

步骤 4：计算第二个积分
计算${\int }_{0}^{1}\dfrac {2}{\sqrt {x+1}}dx$，得到结果。
${\int }_{0}^{1}\dfrac {2}{\sqrt {x+1}}dx=2{\int }_{0}^{1}\dfrac {1}{\sqrt {x+1}}dx=2{\int }_{0}^{1}\dfrac {1}{\sqrt {x+1}}d(x+1)=4\sqrt {x+1}|\dfrac {1}{0}=4\sqrt {2}-4$

步骤 5：合并结果
将两个积分的结果合并，得到最终答案。
${\int }_{0}^{1}\dfrac {x+2}{\sqrt {x+1}}dx=2(\dfrac {2}{3}\sqrt {2}-\dfrac {1}{3}-\sqrt {2}+1)+4\sqrt {2}-4=\dfrac {10\sqrt {2}-8}{3}$