Монгол Бодлогын Сан

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

Бодолт: $\dfrac{1}{(x+1)(x^2+1)}= \dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+1}$ бол $$A(x^2+1)+(Bx+C)(x+1)=1$$ буюу $$(A+B)x^2+(B+C)x+A+C=1$$ тул $$\left\{\begin{array}{c} A+B=0\\ B+C=0\\ A+C=1 \end{array}\right.\Rightarrow\left\{\begin{array}{c} A=\dfrac12\\ B=-\dfrac12\\ C=\dfrac12 \end{array}\right.$$ Иймд \begin{align*} \text{Инт.}&=\int\dfrac{dx}{(x+1)(x^2+1)}\\ &=\int\left(\dfrac{\frac12}{x+1}+\dfrac{-\frac12x+\frac12}{x^2+1}\right)dx\\ &=\dfrac12\int\dfrac{dx}{x+1}-\frac12\int\dfrac{xdx}{x^2+1}+\dfrac12\int\dfrac{dx}{x^2+1}\\ &=\dfrac12\ln|x+1|-\dfrac14\int\dfrac{d(x^2+1)}{x^2+1}\ln|x-2|+\dfrac12\arctg x+C\\ &=\dfrac14\ln(x+1)^2-\dfrac14\ln(x^2+1)+\dfrac12\arctg x+C\\ &=\dfrac14\ln\dfrac{(x+1)^2}{x^2+1}+\dfrac12\arctg x+C \end{align*}