Question

Помогите найти интеграл

Answer 1

$displaystyleintfrac{x^2+1}{x+3}dx=int xdx-3int dx+10intfrac{d(x+3)}{x+3}=\=frac{x^2}{2}-3x+10ln|x+3|+C\\\(x^2+1):(x+3)=x-3+frac{10}{x+3}$

$displaystyle int frac{2x-4}{x^2-3x-4}dx=frac{4}{5}intfrac{d(x-4)}{x-4}+frac{6}{5}intfrac{d(x+1)}{x+1}=\frac{4}{5}ln|x-4|+frac{6}{5}ln|x+1|+C\\\frac{2x-4}{x^2-3x-4}=frac{A}{x-4}+frac{B}{x+1}=frac{4}{5(x-4)}+frac{6}{5(x+1)}\2x-4=A(x+1)+B(x-4)\x|2=A+B\x^0|-4=A-4B\6=5Brightarrow B=frac{6}{5};A=frac{4}{5}$