Question

Помогите найти интеграл

Answer 1

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

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