Cho \(\displaystyle\int\limits_2^3\dfrac{x+2}{2x^2-3x+1}\mathrm{\,d}x=a\ln5+b\ln3+3\ln2\) (\(a,\,b\in\mathbb{Q}\)). Tính \(P=2a-b\).
 | \(P=1\) |
 | \(P=7\) |
 | \(P=-\dfrac{15}{2}\) |
 | \(P=\dfrac{15}{2}\) |
Chọn phương án C.
\(\begin{aligned}\displaystyle\int\limits_2^3\dfrac{x+2}{2x^2-3x+1}\mathrm{\,d}x&=\displaystyle\int\limits_2^3\dfrac{x+2}{(x-1)(2x-1)}\mathrm{\,d}x\\
&=\displaystyle\int\limits_2^3\left(\dfrac{3}{x-1}-\dfrac{5}{2x-1}\right)\mathrm{\,d}x\\
&=\left(3\ln|x-1|-\dfrac{5}{2}\ln|2x-1|\right)\bigg|_2^3\\
&=-\dfrac{5}{2}\ln 5+\dfrac{5}{2}\ln 3+3\ln 2.\end{aligned}\)
Theo đó \(a=-\dfrac{5}{2},\,b=\dfrac{5}{2}\).
Suy ra \(P=2a-b=-\dfrac{15}{2}\).
Giả sử \(\dfrac{x+2}{(x-1)(2x-1)}=\dfrac{A}{x-1}+\dfrac{B}{2x-1}\)
\(\Leftrightarrow\dfrac{x+2}{(x-1)(2x-1)}=\dfrac{(2A+B)x-A-B}{(x-1)(2x-1)}\)
Đồng nhất hệ số ta được $$\begin{cases}2A+B=1\\ -A-B=2\end{cases}\Leftrightarrow\begin{cases}A=3\\ B=-5.\end{cases}$$
Vậy \(\dfrac{x+2}{(x-1)(2x-1)}=\dfrac{3}{x-1}-\dfrac{5}{2x-1}\).