Biết $I=\int\limits_{3}^{4}{\dfrac{dx}{{{x}^{2}}+x}}=a\ln 2+b\ln...

$I=\int\limits_{3}^{4}{\dfrac{dx}{{{x}^{2}}+x}}$. Ta có: $\dfrac{1}{{{x}^{2}}+x}=\dfrac{1}{x\left( x+1 \right)}=\dfrac{1}{x}-\dfrac{1}{x+1}$
Khi đó: $I=\int\limits_{3}^{4}{\dfrac{dx}{{{x}^{2}}+x}}=\int\limits_{3}^{4}{\left( \dfrac{1}{x}-\dfrac{1}{x+1} \right)dx}=\left. \left[ \ln x-\ln \left( x+1 \right) \right] \right|_{3}^{4}$
$=\left( \ln 4-\ln 5 \right)-\left( \ln 3-\ln 4 \right)=4\ln 2-\ln 3-\ln 5$. Suy ra $a=4$, $b=-1$, $c=-1$. Vậy $S=2$
Cách khác: Sử dụng máy tính cần tay hỗ trợ
Ta có: $I=\int\limits_{3}^{4}{\dfrac{dx}{{{x}^{2}}+x}}=a\ln 2+b\ln 3+c\ln 5\Rightarrow {{e}^{t}}={{e}^{a\ln 2+b\ln 3+c\ln 5}}\Rightarrow {{e}^{\ln \left( {{2}^{a}}{{.3}^{b}}{{.5}^{c}} \right)}}$
Hay $\dfrac{16}{15}={{2}^{a}}{{.3}^{b}}{{.5}^{c}}\Leftrightarrow {{2}^{4}}={{2}^{a}}{{.3}^{b+1}}{{.5}^{c+1}}\Rightarrow \left\{ \begin{aligned}
& a=4 \\
& b+1=0 \\
& c+1=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=4 \\
& b=-1 \\
& c=-1 \\
\end{aligned} \right.\Rightarrow S=a+b+c=2$