Cho biết \(\displaystyle\int\limits_0^2\dfrac{x-1}{x^2+4x+3}\mathrm{\,d}x=a\ln5+b\ln3\), với \(a,\,b\in\mathbb{Q}\). Biểu thức \(T=a^2+b^2\) bằng
 | \(13\) |
 | \(10\) |
 | \(25\) |
 | \(5\) |
Chọn phương án A.
\(\begin{aligned}
\displaystyle\int\limits_0^2{\dfrac{x-1}{x^2+4x+3}}\mathrm{\,d}x&=\displaystyle\int\limits_0^2{\dfrac{x-1}{(x+3)(x+1)}}\mathrm{\,d}x\\
&=\displaystyle\int\limits_0^2{\left(\dfrac{2}{x+3}-\dfrac{1}{x+1}\right)}\mathrm{\,d}x\\
&=\left(2\ln|x+3|-\ln|x+1|\right)\bigg|_0^2\\
&=2\ln5-3\ln3.\end{aligned}\)
Theo đó \(a=2,\,b=-3\Rightarrow a^2+b^2=13\).
Giả sử \(\dfrac{x-1}{(x+3)(x+1)}=\dfrac{A}{x+3}+\dfrac{B}{x+1}\)
\(\Leftrightarrow\dfrac{x-1}{(x+3)(x+1)}=\dfrac{(A+B)x+A+3B}{(x+3)(x+1)}\)
Đồng nhất hệ số ta được $$\begin{cases}A+B&=1\\ A+3B&=-1\end{cases}\Leftrightarrow\begin{cases}A=2\\ B=-1.\end{cases}$$
Vậy \(\dfrac{x-1}{(x+3)(x+1)}=\dfrac{2}{x+3}-\dfrac{1}{x+1}\).