∫(x + 2) / (x³ – 2x²) dx
(х+2)/(x2(x–2))=(A/x)+(B/x2)+(D/(x–2))
x+1=A·x·(x–2)+B·(x–2)+D·x2
x+1=(A+D)x2+(B–2A)x–2B
A+D=0
B–2A=1
–2B=1 ⇒ B=–1/2
2A=B–2
2A=–5/2
A=–5/4
D=5/4
Интеграл от суммы равен сумме интегралов:
∫(х+2)dx/(x2(x–2)) = (–5/4) ∫ dx/x –(5/2) ∫ dx/x2+
+(5/4) ∫ dx/(x–2)=
=(–5/4)ln|x|+(5/2)·(1/x)+(5/4)ln|x–2|+C