Question

Помогите: интеграл x^3dx/(x-1)^2(x+3)

Answer 1

$int frac{x^3, dx}{(x-1)^2(x+3)}=int frac{x^3, dx}{(x^2-2x+1)(x+3)}=int frac{x^3}{x^3+x^2-5x+1}dx=\\=int (1+frac{-x^2+5x-1}{x^3+x^2-5x+1})dx=int dx-int frac{x^2-5x+1}{(x-1)^2(x+3)}dx;\\frac{x^2-5x+1}{(x-1)^2(x+3)}=frac{A}{(x-1)^2}+frac{B}{x-1}+frac{C}{x+3}quad to \\x^2-5x+1=A(x+3)+B(x-1)(x+3)+C(x-1)^2; ; to$

$x=1:; ; A=frac{1-5+1}{1+3}=-frac{3}{4}\\x=-3:; ; C=frac{9+15+1}{16}=frac{25}{16}\\x^2|; ; 1=B+C; ; ; rightarrow ; ; ; B=C-1=frac{25}{16}-1=frac{9}{16}$

$int frac{x^3, dx}{(x-1)^2(x+3)}=x+frac{3}{4}int frac{dx}{(x-1)^2}-frac{9}{16}int frac{dx}{x-1}-frac{25}{16}int frac{dx}{x+3}=\\=x+frac{3}{4}cdot frac{(x-1)^3}{3}-frac{25}{16}cdot ln|x+3|+C$