Đáp án:
\[F\left( 2 \right) = \frac{4}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
F\left( x \right) = \int {f\left( x \right)dx} = \int {\frac{{2x + 1}}{{{{\left( {{x^2} + x} \right)}^2}}}dx} \\
t = {x^2} + x \Rightarrow dt = \left( {{x^2} + x} \right)'dx = \left( {2x + 1} \right)dx\\
\Rightarrow F\left( x \right) = \int {\frac{{dt}}{{{t^2}}}} = - \frac{1}{t} + C = - \frac{1}{{{x^2} + x}} + C\\
F\left( 1 \right) = 1 \Leftrightarrow - \frac{1}{{{1^2} + 1}} + C = 1 \Leftrightarrow C = \frac{3}{2}\\
\Rightarrow F\left( x \right) = - \frac{1}{{{x^2} + x}} + \frac{3}{2}\\
\Rightarrow F\left( 2 \right) = - \frac{1}{{{2^2} + 2}} + \frac{3}{2} = \frac{4}{3}
\end{array}\)
Vậy \(F\left( 2 \right) = \frac{4}{3}\)
