Question

Найти неопределенный интеграл: ∫dx/x⁴-x²

Answer 1

Пошаговое объяснение:

$\displaystyle \int\frac{dx}{x^4-x^2}=\int\frac{dx}{x^2(x-1)(x+1)}=\int\left(\frac{A}{x^2}+\frac{B}{x+1}+\frac{C}{x-1}\right)dx~~\boxed{=}\\ \\ \\ \frac{1}{x^2(x-1)(x+1)}=\frac{A(x+1)(x-1)+Bx^2(x-1)+Cx^2(x+1)}{x^2(x-1)(x+1)}\\ \\ \\ x^{-1}:~~~~ 1=-2B~~~~\Longleftrightarrow~~~~~B=-\frac{1}{2}\\ x^1~~:~~~1=2C~~~~\Longleftrightarrow~~~~ C=\frac{1}{2}\\ \\ x^0~~:~~~1=-A~~~~\Longleftrightarrow~~~~A=-1$

$\displaystyle \boxed{=}~\int -\frac{dx}{x^2}-\frac{1}{2}\int\frac{dx}{x+1}+\frac{1}{2}\int\frac{dx}{x-1}=\frac{1}{x}-\frac{1}{2}\ln|x+1|+\frac{1}{2}\ln|x-1|+C=\\ \\ \\ =\frac{1}{x}-\frac{1}{2}\ln\bigg|\frac{x+1}{x-1}\bigg|+C$