Biết \(\displaystyle\int\limits_1^3\dfrac{\mathrm{\,d}x}{\sqrt{x+1}-\sqrt{x}}=a\sqrt{3}+b\sqrt{2}+c\) với \(a\), \(b\), \(c\) là các số hữu tỷ. Tính \(P =a+b+c\).
 | \(P=\dfrac{13}{2}\) |
 | \(P=\dfrac{16}{3}\) |
 | \(P=5\) |
 | \(P=\dfrac{2}{3}\) |
Chọn phương án B.
\(\begin{aligned}
&\,\displaystyle\int\limits_1^3\dfrac{\mathrm{\,d}x}{\sqrt{x+1}-\sqrt{x}}\\
=&\,\displaystyle\int\limits_1^3\dfrac{\sqrt{x+1}+\sqrt{x}}{\left(\sqrt{x+1}-\sqrt{x}\right)\left(\sqrt{x+1}+\sqrt{x}\right)}\mathrm{\,d}x\\ =&\,\displaystyle\int\limits_1^3\dfrac{\sqrt{x+1}+\sqrt{x}}{(x+1)-1}\mathrm{\,d}x\\
=&\,\displaystyle\int\limits_1^3\left(\sqrt{x+1}+\sqrt{x}\right)\mathrm{\,d}x\\
=&\,\displaystyle\int\limits_1^3\left((x+1)^{\frac{1}{2}}+x^{\frac{1}{2}}\right)\mathrm{\,d}x\\
=&\,\left[\dfrac{2}{3}\left(\sqrt{x+1}\right)^3 +\dfrac{2}{3}\left(\sqrt{x}\right)^3\right]\bigg|_1^3\\
=&\,\dfrac{16}{3}+2\sqrt{3}-\dfrac{4}{3}\sqrt{2}-\dfrac{2}{3}\\
=&\,2\sqrt{3}-\dfrac{4}{3}\sqrt{2}+\dfrac{14}{3}.\end{aligned}\)
Theo đó \(a=2,\,b=-\dfrac{4}{3},\,c=\dfrac{14}{3}\).
\(\Rightarrow P=a+b+c=2-\dfrac{4}{3}+\dfrac{14}{3}=\dfrac{16}{3}.\)