Монгол Бодлогын Сан

Заавар: $x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)$ тул $$\dfrac{1}{x^4-1}=\dfrac{A}{x-1}+\dfrac{B}{x+1}+\dfrac{Cx+D}{x^2+1}$$ байх $A$, $B$, $C$, $D$ тоонуудыг ол.

Бодолт: $\dfrac{1}{x^4-1}=\dfrac{A}{x-1}+\dfrac{B}{x+1}+\dfrac{Cx+D}{x^2+1}$ бол $$A(x^2+1)(x+1)+B(x^2+1)(x-1)+(Cx+D)(x^2-1)=1$$ буюу $$(A+B+C)x^3+(A-B+D)x^2+(A+B-C)x+A-B-D=1$$ тул $$\left\{\begin{array}{c} A+B+C=0\\ A-B+D=0\\ A+B-C=0\\ A-B-D=1 \end{array}\right.\Rightarrow\left\{\begin{array}{l} A=\dfrac14\\ B=-\dfrac14\\ C=0\\ D=-\dfrac12 \end{array}\right.$$ Иймд \begin{align*} \text{Инт.}&=\int\dfrac{dx}{x^4-1}=\int\left(\dfrac{\frac14}{x-1}+\dfrac{-\frac14}{x+1}+\dfrac{-\frac12}{x^2+1}\right)dx\\ &=\dfrac14\int\dfrac{dx}{x-1}-\frac14\int\dfrac{dx}{x+1}-\dfrac12\int\dfrac{dx}{x^2+1}dx\\ &=\dfrac14\ln|x-1|-\dfrac14\ln|x+1|-\frac12\arctg x+C\\ &=\dfrac14\ln\left|\dfrac{x-1}{x+1}\right|-\frac12\arctg x+C \end{align*}