Biết $I=\displaystyle\displaystyle\int\limits_{1}^{2}\dfrac{\mathrm{d}x}{(x+1)\sqrt{x}+x\sqrt{x+1}}=\sqrt{a}-\sqrt{b}-c$ với $a,\,b,\,c$ là các số nguyên dương. Tính $P=a+b+c$.
 | $P=18$ |
 | $P=12$ |
 | $P=24$ |
 | $P=46$ |
Chọn phương án D.
$\begin{aligned}
I&=\displaystyle\int\limits_{1}^{2}\dfrac{\mathrm{d}x}{\sqrt{x}\cdot\sqrt{x+1}\left(\sqrt{x}+\sqrt{x+1}\right)}\\
&=\displaystyle\int\limits_{1}^{2}\dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}\cdot\sqrt{x+1}\left(\sqrt{x+1}+\sqrt{x}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}\mathrm{\,d}x\\
&=\displaystyle\int\limits_{1}^{2}\dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}\cdot\sqrt{x+1}\left((x+1)-x\right)}\mathrm{\,d}x\\
&=\displaystyle\int\limits_{1}^{2}\dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}\cdot\sqrt{x+1}}\mathrm{\,d}x\\
&=\displaystyle\int\limits_{1}^{2}\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+1}}\right)\mathrm{\,d}x\\
&=\displaystyle\int\limits_{1}^{2}\left(\dfrac{2}{2\sqrt{x}}-\dfrac{2}{2\sqrt{x+1}}\right)\mathrm{\,d}x\\
&=\left(2\sqrt{x}-2\sqrt{x+1}\right)\bigg|_1^2\\
&=4\sqrt{2}-2\sqrt{3}-2=\sqrt{32}-\sqrt{12}-2.
\end{aligned}$
Suy ra $a=32$, $b=12$, $c=2$. Vậy $P=46$.