Question

(5x^4-x^3+4x^2+8)/(x^3-8) интеграл

Answer 1

$int frac{5x^4-x^3+4x^2+8}{x^3-8}=int (5x-1+frac{4x^2+40x}{(x-2)(x^2+2x+4)})dx=\\=frac{5x^2}{2}-x+int frac{2, dx}{x-2}+int frac{-4x+16}{x^2+2x+4}dx=2,5x^2-x+2ln|x-2|-\\-4int frac{(x-4)dx}{(x+1)^2+3}=2,5x^2-x+2ln|x-2|-4int frac{(t-5)dt}{t^2+3}=[t=x+1]=\\=2,5x^2-x+2ln|x-2|-2int frac{2t, dt}{t^2+3}+20int frac{dt}{t^2+3}=[2t, dt=d(t^2+3)]=\\=2,5x^2-x+2ln|x-2|-2ln|t^2+3|+frac{20}{sqrt3}cdot arctgfrac{t}{sqrt3}+C=\\=2,5x^2-x+2ln|x-2|-2ln|x^2+2x+4|+frac{20}{sqrt3}cdot arctgfrac{x+1}{sqrt3}+C$

$P.S.; ; frac{4x^2+40x}{(x-2)(x^2+2x+4)}=frac{A}{x-2}+frac{Bx+C}{x^2+2x+4}\\4x^2+40x=Ax^2+2Ax+4A+Bx^2-2Bx+Cx-2C\\x=2:; ; A=frac{96}{12}=8\\x^2|; 4=A+B; ,; ; B=4-A=-4\x^0|; 0=4A-2C; ,; ; C=2A=16\\frac{4x^2+40x}{(x-2)(x^2+2x+4)}=frac{8}{x-2}+frac{-4x+16}{x^2+2x+4}$

Answer 2

а спасибо?

Answer 3

❤️

Answer 4

кнопочка "спасибо"