c: Đặt \(\frac{x}{x^2-5x+6}=\frac{x}{\left(x-2\right)\left(x-3\right)}=\frac{A}{x-2}+\frac{B}{x-3}\)
=>\(\frac{x}{\left(x-2\right)\left(x-3\right)}=\frac{A\left(x-3\right)+B\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
=>x=A(x-3)+B(x-2)=x(A+B)-3A-2B
=>A+B=1 và -3A-2B=0
=>A+B=1 và 3A+2B=0
=>\(\begin{cases}3A+3B=3\\ 3A+2B=0\end{cases}\Rightarrow\begin{cases}3A+3B-3A-2B=3-0=3\\ A+B=1\end{cases}\)
=>\(\begin{cases}B=3\\ A=1-3=-2\end{cases}\)
=>\(\int\frac{\mathrm{x}}{x^2-5x+6}\mathrm{d}x=\int_{}^{}\!-\frac{2}{x-2}+\frac{3}{x-3}\,\mathrm{d}x\)
\(=-2\cdot\ln\left|x-2\right|+3\cdot\ln\left|x-3\right|+C\)
d: Đặt \(\frac{3x+2}{x^2-1}=\frac{3x+2}{\left(x+1\right)\left(x-1\right)}=\frac{A}{x+1}+\frac{B}{x-1}\)
=>\(\frac{A\left(x-1\right)+B\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{3x+2}{\left(x-1\right)\left(x+1\right)}\)
=>A(x-1)+B(x+1)=3x+2
=>x(A+B)-A+B=3x+2
=>A+B=3 và -A+B=2
=>A+B-A+B=3+2=5
=>2B=5
=>B=2,5
A+B=3
=>A=3-2,5=0,5
\(\int_{}^{}\!\frac{3x+2}{x^2-1}\,\mathrm{d}x=\int_{}^{}\!\frac{3x+2}{\left(x-1\right)\left(x+1\right)}\,\mathrm{d}x\)
\(\)\(=\int_{}^{}\frac{0.5}{x+1}+\frac{2.5}{x-1}\!\mathrm{d}x=0,5\cdot\ln\left|x+1\right|+2,5\cdot\ln\left|x-1\right|\) +C
