Biết \(\int_1^2\dfrac{\text{d}x}{\left(x+1\right)\sqrt{x}+x\sqrt{x+1}}=\sqrt{a}-\sqrt{b}-c\) với \(a,b,c\) là các số nguyên dương. Tổng \(a+b+c\) bằng
\(24\).\(12\).\(18\).\(46\).Hướng dẫn giải:Ta có
\(\dfrac{1}{\left(x+1\right)\sqrt{x}+x\sqrt{x+1}}=\dfrac{1}{\sqrt{x\left(x+1\right)}\left(\sqrt{x+1}+\sqrt{x}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}.\sqrt{x+1}}=\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+1}}\)
Vậy thì \(\int_1^2\dfrac{\text{d}x}{\left(x+1\right)\sqrt{x}+x\sqrt{x+1}}=\int_1^2\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+1}}\right)\text{d}x=\int_1^2\dfrac{1}{\sqrt{x}}\text{d}x-\int_1^2\dfrac{1}{\sqrt{x+1}}\text{d}x\)\(=\int_1^2x^{-\frac{1}{2}}\text{d}x-\int_1^2\left(x+1\right)^{-\frac{1}{2}}\text{d}x\)
\(=2\left(\sqrt{x}-\sqrt{x+1}\right)|^2_1=2\sqrt{2}-2\sqrt{3}-2+2\sqrt{2}=4\sqrt{2}-2\sqrt{3}-2=\sqrt{32}-\sqrt{12}-2\)
Do đó \(a=32,b=12,c=2,a+b+c=46.\)
