Монгол Бодлогын Сан

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

Бодолт: $\dfrac{3x+1}{x(x+1)(x-2)}=\dfrac{A}{x}+\dfrac{B}{x+1}+\dfrac{C}{x-2}$ бол $$3x+1=A(x+1)(x-2)+Bx(x-2)+Cx(x+1)$$ байна. $x=0,-1,2$ утгууд дээр бодвол $1=-2A$, $-2=3B$, $7=6C$ тул $$\dfrac{3x+1}{x(x+1)(x-2)}=\dfrac{-\frac12}{x}+\dfrac{-\frac23}{x+1}+\dfrac{\frac76}{x-2}$$ болно. Иймд \begin{align*} \int \dfrac{3x+1}{x(x+1)(x-2)}dx&=\int\left(\dfrac{-\frac12}{x}+\dfrac{-\frac23}{x+1}+\dfrac{\frac76}{x-2}\right)dx\\ &=-\dfrac12\int\dfrac{1}{x}-\dfrac23\int\dfrac{1}{x+1}+\dfrac76\int\dfrac{1}{x-2}+C\\ &=-\dfrac12\ln|x|-\dfrac{2}{3}\ln|x+1|+\dfrac76\ln|x-2|+C\\ &=-\dfrac16\ln|x^3|-\dfrac{1}{6}\ln(x+1)^4+\dfrac16\ln|x-2|^7+C\\ &=\dfrac16\ln\left|\dfrac{(x-2)^7}{x^3(x+1)^4}\right|+C \end{align*}