Biết \(\displaystyle\int\limits_1^2\dfrac{4\mathrm{\,d}x}{(x+4)\sqrt{x}+x\sqrt{x+4}}=\sqrt{a}+\sqrt{b}-\sqrt{c}-d\) với \(a\), \(b\), \(c\), \(d\) là các số nguyên dương. Tính \(P=a+b+c+d\).
 | \(48\) |
 | \(46\) |
 | \(54\) |
 | \(52\) |
Chọn phương án C.
\(\begin{aligned}
I&=\displaystyle\int\limits_1^2\dfrac{4\mathrm{\,d}x}{(x+4)\sqrt{x}+x\sqrt{x+4}}\\
&=\displaystyle\int\limits_1^2\dfrac{4}{\sqrt{x(x+4)}\left(\sqrt{x+4}+\sqrt{x}\right)}\mathrm{\,d}x\\
&=\displaystyle\int\limits_1^2\dfrac{4\left(\sqrt{x+4}-\sqrt{x}\right)}{\sqrt{x(x+4)}\left(\sqrt{x+4}+\sqrt{x}\right)\left(\sqrt{x+4}-\sqrt{x}\right)}\mathrm{\,d}x\\
&=\displaystyle\int\limits_{1}^2\dfrac{4\left(\sqrt{x+4}-\sqrt{x}\right)}{\sqrt{x(x+4)}\left((x+4)-x\right)}\mathrm{\,d}x\\
&=\displaystyle\int\limits_{1}^2\dfrac{\sqrt{x+4}-\sqrt{x}}{\sqrt{x(x+4)}}\mathrm{\,d}x\\
&=\displaystyle\int\limits_{1}^2\dfrac{\sqrt{x+4}-\sqrt{x}}{\sqrt{x}\cdot\sqrt{x+4}}\mathrm{\,d}x\\
&=\displaystyle\int\limits_{1}^2\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+4}}\right)\mathrm{\,d}x\\
&=\displaystyle\int\limits_{1}^2\left(\dfrac{2}{2\sqrt{x}}-\dfrac{2}{2\sqrt{x+4}}\right)\mathrm{\,d}x\\
&=\left(2\sqrt{x}-2\sqrt{x+4}\right)\bigg|_1^2\\
&=2\sqrt{2}-2\sqrt{6}-2+2\sqrt{5}\\
&=\sqrt{8}+\sqrt{20}-\sqrt{24}-2.\end{aligned}\)
Theo đó \(a=8,\,b=20,\,c=24,\,d=2\).
Suy ra \(P=a+b+c+d=54\).