Đáp án: $A$
Giải thích các bước giải:
Ta có:
$f'(x)=\dfrac{2x^3-\sqrt{x+1}}{x^2\sqrt{x+1}}$
$\to f'(x)=\dfrac{2x}{\sqrt{x+1}} -\dfrac{1}{x^2}$
$\to f'(x)=\dfrac{2(x+1)- 2}{\sqrt{x+1}} -\dfrac{1}{x^2}$
$\to f'(x)=2\sqrt{x+1}-\dfrac{2}{\sqrt{x+1}} -\dfrac{1}{x^2}$
$\to f'(x)=2(x+1)^{\frac12}-2(x+1)^{-\frac12} -\dfrac{1}{x^2}$
$\to f(x)=\displaystyle\int f'(x)dx=\displaystyle\int 2(x+1)^{\frac12}-2(x+1)^{-\frac12} -\dfrac{1}{x^2}dx$
$\to f(x)=\displaystyle\int 2(x+1)^{\frac12}-2(x+1)^{-\frac12} -\dfrac{1}{x^2}dx$
$\to f(x)=\dfrac{4}{3}\left(x+1\right)^{\frac{3}{2}}-4\left(x+1\right)^{\frac{1}{2}}+\dfrac{1}{x}+C$
Vì $f(3)=6$
$\to \dfrac{4}{3}\left(3+1\right)^{\frac{3}{2}}-4\left(3+1\right)^{\frac{1}{2}}+\dfrac{1}{3}+C=6$
$\to C=3$
$\to f(x)=\dfrac{4}{3}\left(x+1\right)^{\frac{3}{2}}-4\left(x+1\right)^{\frac{1}{2}}+\dfrac{1}{x}+3$
$\to f(8)=\dfrac{4}{3}\left(8+1\right)^{\frac{3}{2}}-4\left(8+1\right)^{\frac{1}{2}}+\dfrac{1}{8}+3$
$\to f(8)=\dfrac{217}{8}$
