Монгол Бодлогын Сан

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

Бодолт: $\dfrac{1}{(x+1)(x^2-x+1)}= \dfrac{A}{x+1}+\dfrac{Bx+C}{x^2-x+1}$ бол $$A(x^2-x+1)+(Bx+C)(x+1)=1$$ буюу $$(A+B)x^2+(-A+B+C)x+A+C=1$$ тул $$\left\{\begin{array}{c} A+B=0\\ -A+B+C=0\\ A+C=1 \end{array}\right.\Rightarrow\left\{\begin{array}{c} A=\dfrac13\\ B=-\dfrac13\\ C=\dfrac23 \end{array}\right.$$ Иймд $$\begin{align*} \text{Инт.}&=\int\dfrac{dx}{x^3+1}=\int\dfrac{dx}{(x+1)(x^2-x+1)}\\ &=\int\left(\dfrac{\frac13}{x+1}+\dfrac{-\frac13x+\frac23}{x^2-x+1}\right)dx\\ &=\dfrac13\int\dfrac{dx}{x+1}-\frac13\int\dfrac{x-\frac12-\frac32}{x^2-x+1}dx\\ &=\dfrac13\ln|x+1|-\dfrac13\int\dfrac{x-\frac12}{x^2-x+1}dx+\frac12\int\dfrac{dx}{x^2-x+1}\\ &=\dfrac13\ln|x+1|-\dfrac16\int\dfrac{(2x-1)dx}{x^2-x+1}+\frac12\int\dfrac{dx}{\left(x-\frac12\right)^2+\frac34}\\ &=\dfrac13\ln|x+1|-\dfrac16\int\dfrac{d(x^2-x+1)}{x^2-x+1}+\frac12\int\dfrac{d\left(x-\frac12\right)}{\left(x-\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}\\ &=\dfrac16\ln(x+1)^2-\dfrac16\ln(x^2-x+1)+\dfrac1{\sqrt3}\arctg\dfrac{2x-1}{\sqrt3}+C\\ &=\dfrac16\ln\dfrac{(x+1)^2}{x^2-x+1}+\dfrac{1}{\sqrt3}\arctg\dfrac{2x-1}{\sqrt3}+C \end{align*}$$