Cho hàm số f(x) có f(3) = 3 và \(f(x)=\frac{x}{x+1-\sqrt{x+1}},\forall x>0\). Khi đó \(\int\limits_{3}^{8}{f(x)dx}\) bằng

* Hướng dẫn giải

Ta có \(f\left( x \right)=\int{f'\left( x \right)\text{dx}}=\int{\frac{x}{x+1-\sqrt{x+1}}\text{dx}}\)

                                                    \(=\int{\frac{x\left( x+1+\sqrt{x+1} \right)}{{{\left( x+1 \right)}^{2}}-\left( x+1 \right)}\text{dx=}}\int{\left( \text{1+}\frac{1}{\sqrt{x+1}} \right)\text{dx}}=x+2\sqrt{x+1}+C\)

Ta có \(f\left( 3 \right)=3\Leftrightarrow C=-4\) suy ra \(f\left( x \right)=x+2\sqrt{x+1}-4\)

Khi đó \(\int\limits_{3}^{8}{f\left( x \right)\text{d}x}=\int\limits_{3}^{8}{\left( x+2\sqrt{x+1}-4 \right)\text{d}x}=\frac{197}{6}\)