Монгол Бодлогын Сан

Заавар: $\dfrac{x^2}{(x-1)(x-2)(x-3)}= \dfrac{A}{x-1}+\dfrac{B}{x-2}+\dfrac{C}{x-3}$ байх $A$, $B$, $C$ тоонуудыг ол.

Бодолт: $\dfrac{x^2}{(x-1)(x-2)(x-3)}= \dfrac{A}{x-1}+\dfrac{B}{x-2}+\dfrac{C}{x-3}$ бол $$A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)=x^2$$ буюу $$(A+B+C)x^2-(5A+4B+3C)x+6A+3B+2C=x^2$$ тул $$\left\{\begin{array}{c} A+B+C=1\\ 5A+4B+3C=0\\ 6A+3B+2C=0 \end{array}\right.\Rightarrow\left\{\begin{array}{c} A=\dfrac12\\ B=-4\\ C=\dfrac92 \end{array}\right.$$ Иймд $$\begin{align*} \text{Инт.}&=\int\dfrac{x^2dx}{(x-1)(x-2)(x-3)}\\ &=\int\left(\dfrac{\frac12}{x-1}-\dfrac{4}{x-2}+\dfrac{\frac92}{x-3}\right)dx\\ &=\dfrac12\int\dfrac{dx}{x-1}-4\int\dfrac{dx}{x-2}+\dfrac92\int\dfrac{dx}{x-3}\\ &=\dfrac12\ln|x-1|-4\ln|x-2|+\dfrac92\ln|x-3|+C \end{align*}$$