Tính \(J = \mathop \smallint \nolimits_2^3 \frac{{2x + 3}}{{{x^3} - 3x + 2}}dx\)
Suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo saiTa có: \({x^3} - 3x + 2 = {\left( {x - 1} \right)^2}\left( {x + 2} \right)\)
\(\begin{array}{l}
2x + 3 = a{\left( {x - 1} \right)^2} + b\left( {x + 2} \right)\left( {x - 1} \right) + c\left( {x + 2} \right)\\
\Leftrightarrow 2x + 3 = \left( {a + b} \right){x^2} + \left( {c - 2a + b} \right)x + a - 2b + 2c
\end{array}\)
\(\begin{array}{l}
\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{a + b = 0}\\
{ - 2a + b + c = 2}\\
{a - 2b + 2c = 3}
\end{array}} \right.\\
\Leftrightarrow a = - \frac{1}{9},b = \frac{1}{9},c = \frac{5}{3}
\end{array}\)
\(\begin{array}{l}
J = \mathop \smallint \nolimits_2^3 \left[ { - \frac{1}{9}\frac{1}{{x + 2}} + \frac{1}{9}\frac{1}{{x - 1}} + \frac{5}{3}\frac{1}{{{{\left( {x - 1} \right)}^2}}}} \right]dx\\
\; = \left( {\frac{1}{9}\ln \left| {\frac{{x - 1}}{{x + 2}}} \right| - \frac{5}{{3\left( {x - 1} \right)}}} \right)|_2^3 = \frac{1}{9}\ln \frac{8}{5} + \frac{5}{6}
\end{array}\)