Question

Помогите решить интеграл пожалуйста

Answer 1

$\displaystyle \int \frac{dx}{x(x+4)}=\int(\frac{1}{4x}-\frac{1}{4(x+4)})dx=\frac{1}{4}(\int\frac{dx}{x}-\int\frac{dx}{x+4})=\\=\frac{1}{4}(ln|x|-ln|x+4|)+C=\frac{1}{4}ln|\frac{x}{x+4}|+C\\\\\\\frac{1}{x(x+4)}=\frac{A}{x}+\frac{B}{x+4}=\frac{1}{4x}-\frac{1}{4(x+4)}\\1=A(x+4)+Bx\\x|0=A+B=\ \textgreater \ B=-\frac{1}{4}\\x^0|1=4A=\ \textgreater \ A=\frac{1}{4}$