[imath]\dfrac{x+3}{x^2+3x+2} = \dfrac{x+3}{(x+1)(x+2)} \\
= \dfrac{A}{x+1} + \dfrac{B}{x+2} \\
= \dfrac{A(x+2) + B(x+1)}{(x+1)(x+2)} \\
= \dfrac{(A+B)x + 2A+B}{(x+1)(x+2)}[/imath]
Ta có hệ: [imath]
\left\{\begin{matrix}
A+B=1 \\
2A+B=3
\end{matrix}\right. \\
\Leftrightarrow \left\{\begin{matrix}
A=2 \\
B=-1
\end{matrix}\right.
[/imath]
Nên [imath]
\displaystyle \int_{1}^{3}{\dfrac{x+3}{x^2+3x+2}} dx = \int_{1}^{3}{\left (\dfrac{2}{x+1} - \dfrac{1}{x+2} \right )}dx \\
= \int_{1}^{3}{\dfrac{2}{x+1}}dx - \int_{1}^{3}{\dfrac{1}{x+2}}dx \\
\displaystyle = 2 (\ln|x+1|)|^{3}_{1} - (\ln|x+2|)|^{3}_{1} \\
= 2\ln 4 -2 \ln 2 - ( \ln 5 - \ln 3) \\
=2 \ln 2 + \ln 3 - \ln 5 [/imath]
Do đó [imath] a=2, \ b=1, \ c = -1
\Rightarrow a+b+c=2[/imath]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
