Question

Пожалуйста, помогите решить интеграл (ы)

Answer 1

$\displaystyle \int\frac{dx}{x^2+10x+16}=\frac{1}{6}\int\frac{d(x+2)}{x+2}-\frac{1}{6}\int\frac{d(x+8)}{x+8}=\\\frac{1}{6}ln|x+2|-\frac{1}{6}ln|x+8|+C=\frac{1}{6}ln|\frac{x+2}{x+8}|+C\\\\\\\frac{1}{x^2+10x+16}=\frac{A}{x+2}+\frac{B}{x+8}=\frac{1}{6(x+2)}-\frac{1}{6(x+8)}\\1=A(x+8)+B(x+2)\\x|0=A+B\\x^0|1=8A+2B\\-1=-6A\rightarrow A=\frac{1}{6};B=-\frac{1}{6}$

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