Монгол Бодлогын Сан

Заавар: $\dfrac{2x + 3}{x^2-9}= \dfrac{2x + 3}{(x- 3)(x + 3)}=\dfrac{A}{x - 3} + \dfrac{B}{x + 3}$ байх $A$, $B$ тоонуудыг ол.

Бодолт: $\dfrac{2x + 3}{x^2-9}= \dfrac{2x + 3}{(x- 3)(x + 3)}=\dfrac{A}{x - 3} + \dfrac{B}{x + 3}$-ээс $$A(x+3)+B(x-3)=2x+3\Rightarrow Ax+3A+Bx-3B=2x+3$$ тул $$(A+B)x+3A-3B=2x+3$$ буюу $$\left\{\begin{array}{c} A+B=2\\ 3A-3B=3 \end{array}\right.\Rightarrow\left\{\begin{array}{c} A=\dfrac32\\ B=\dfrac12 \end{array}\right.$$ Иймд $$\dfrac{2x+3}{x^2-9}=\dfrac{\frac32}{x-3}+\dfrac{\frac12}{x+3}$$ болох ба интеграл нь $$\int\dfrac{2x+3}{x^2-9}dx=\dfrac32\int\dfrac{dx}{x-3}+\dfrac12\int\dfrac{dx}{x+3}=\dfrac32\ln|x-3|+\dfrac12\ln|x+3|+C$$