第7章 7.3

(1)

$\frac{5}{(2\text{x}-1)(\text{x}+2)} = \frac{\text{a}}{2\text{x}-1} + \frac{\text{b}}{\text{x}+2} = \frac{\text{a}(\text{x}+2)+\text{b}(2\text{x}-1)}{(2\text{x}-1)(\text{x}+2)} = \frac{(a+2b)x+2a-b}{(2x-1)(x+2)}$

これから

$a+2b=0 \quad 2a-b=5 \quad \text{となればよいので} \quad a=2 \quad b=-1$

$\frac{2}{2\text{x}-1} - \frac{1}{\text{x}+2}$

(2)

$\frac{4(\text{x}+2)}{(\text{x}+1)^2(\text{x}+3)} = \frac{a}{(x+1)^2} + \frac{b}{x+1} + \frac{c}{x+3} = \frac{a(x+3)+b(x+1)(x+3)+c(x+1)^2}{(x+1)^2(x+3)}$

$= \frac{\text{a}(\text{x}+3)+\text{b}(\text{x}^2+4\text{x}+3)+\text{c}(\text{x}^2+2\text{x}+1)}{(\text{x}+1)^2(\text{x}+3)} = \frac{(b+c)x^2+(a+4b+2c)x+3a+3b+c}{(x+1)^2(x+3)}$

これから

$b+c=0 \quad a+4b+2c=4 \quad 3a+3b+c=8 \quad \text{となればよいので} \quad a=2 \quad b=1 \quad c=-1$

$\frac{2}{(x+1)^2} + \frac{1}{x+1} - \frac{1}{x+3}$