Biết rằng \(\displaystyle\int\limits_2^3 \dfrac{5x+12}{x^2+5x+6}\mathrm{\,d}x=a\ln2+b\ln5+c\ln6\). Tính \(S=3a+2b+c\).
 | \(-11\) |
 | \(-14\) |
 | \(-2\) |
 | \(3\) |
Chọn phương án A.
\(\begin{eqnarray*}
&&\displaystyle\int\limits_2^3\dfrac{5x+12}{x^2+5x+6}\mathrm{\,d}x\\
&=&\displaystyle\int\limits_2^3\dfrac{5x+12}{(x+3)(x+2)}\mathrm{\,d}x\\
&=&\displaystyle\int\limits_2^3\left(\dfrac{3}{x+3}+\dfrac{2}{x+2}\right)\mathrm{\,d}x\\
&=&-4\ln2-\ln5+3\ln6.
\end{eqnarray*}\)
Vậy \(a=-4,\,b=-1,\,c=3\).
Suy ra \(3a+2b+c=-12-2+3=-11\).
Giả sử \(\dfrac{5x+12}{(x+3)(x+2)}=\dfrac{A}{x+3}+\dfrac{B}{x+2}\)
\(\Leftrightarrow\dfrac{5x+12}{(x+3)(x+2)}=\dfrac{(A+B)x+2A+3B}{(x+3)(x+2)}\)
Đồng nhất hệ số ta được $$\begin{cases}A+B&=5\\ 2A+3B&=12\end{cases}\Leftrightarrow\begin{cases}A=3\\ B=2.\end{cases}$$
Vậy \(\dfrac{5x+12}{(x+3)(x+2)}=\dfrac{3}{x+3}+\dfrac{2}{x+2}\).