Chọn phương án D.

\(\begin{eqnarray*}
&I&=\dfrac{1}{2}\displaystyle\int\limits_1^2\dfrac{\mathrm{\,d}x}{(x+1)\sqrt{x}+x\sqrt{x+1}}\\
&&=\dfrac{1}{2}\displaystyle\int\limits_1^2\dfrac{\mathrm{\,d}x}{\sqrt{x+1}\cdot\sqrt{x}\left(\sqrt{x+1}+\sqrt{x}\right)}\\
&&=\dfrac{1}{2}\displaystyle\int\limits_1^2\dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}\cdot\sqrt{x}\left(\sqrt{x+1}+\sqrt{x}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}\mathrm{\,d}x\\
&&=\dfrac{1}{2}\displaystyle\int\limits_1^2\dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}\cdot\sqrt{x}\left((x+1)-x\right)}\mathrm{\,d}x\\
&&=\dfrac{1}{2}\displaystyle\int\limits_1^2 \dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}\cdot\sqrt{x}}\mathrm{\,d}x\\
&&=\dfrac{1}{2}\displaystyle\int\limits_1^2 \left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+1}}\right)\mathrm{\,d}x\\
&&=\displaystyle\int\limits_1^2 \left(\dfrac{1}{2\sqrt{x}}-\dfrac{1}{2\sqrt{x+1}}\right)\mathrm{\,d}x\\
&&=\left(\sqrt{x}-\sqrt{x+1}\right)\bigg|_1^2\\
&&=2\sqrt{2}-\sqrt{3}-1\\
&&=\dfrac{4\sqrt{2}-2\sqrt{3}-2}{2}\\
&&=\dfrac{\sqrt{32}-\sqrt{12}-2}{2}.\end{eqnarray*}\)

Theo đó \(a=32,\,b=12,\,c=2\).
\(\Rightarrow P=a-b+c=32-12+2=22\).