Đáp án:
\[I = - \frac{1}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f'\left( x \right) = - \left( {2x + 1} \right).{\left[ {f\left( x \right)} \right]^2}\\
\Leftrightarrow - \frac{{f'\left( x \right)}}{{{{\left[ {f\left( x \right)} \right]}^2}}} = 2x + 1\\
\Leftrightarrow \int {\frac{{ - f'\left( x \right)}}{{{{\left[ {f\left( x \right)} \right]}^2}}}dx = \int {\left( {2x + 1} \right)dx} } \\
\Leftrightarrow \frac{1}{{f\left( x \right)}} = {x^2} + x + C\\
\Leftrightarrow f\left( x \right) = \frac{1}{{{x^2} + x + C}}\\
f\left( 0 \right) = 1 \Rightarrow \frac{1}{{{0^2} + 0 + C}} \Rightarrow C = 1\\
\Rightarrow f\left( x \right) = \frac{1}{{{x^2} + x + 1}}\\
I = \int\limits_0^1 {\left( {{x^3} - 1} \right)f\left( x \right)dx} \\
= \int\limits_0^1 {\left( {x - 1} \right).\left( {{x^2} + x + 1} \right).\frac{1}{{{x^2} + x + 1}}dx} \\
= \int\limits_0^1 {\left( {x - 1} \right)dx} \\
= \mathop {\left. {\left( {\frac{{{x^2}}}{2} - x} \right)} \right|}\nolimits_0^1 \\
= \left( {\frac{{{1^2}}}{2} - 1} \right) - \left( {\frac{{{0^2}}}{2} - 0} \right)\\
= - \frac{1}{2}
\end{array}\)
