Question

Нужно найти неопределенный интеграл.​

Answer 1

$\int \frac{dx}{2x^2+3x}=\int \frac{dx}{x(2x+3)}=\int \Big (\frac{1}{3x}-\frac{2}{3(2x+3)}\Big )dx=\frac{1}{3}ln|x|-\frac{2}{3\cdot 2}ln|2x+3|+C=\\\\=\frac{1}{3}\cdot (ln|x|-ln|2x+3|)+C=\frac{1}{3}\cdot ln\Big |\frac{x}{2x+3}\Big |+C\; ;\\\\\\\frac{1}{x(2x+3)}=\frac{A}{x}+\frac{B}{2x+3}=\frac{A(2x+3)+Bx}{x(2x+3)}\\\\A(2x+3)+Bx=1\\\\x=0:\; \; A=\frac{1}{2x+3}=\frac{1}{3}\\\\x=-\frac{3}{2}:\; \; B=\frac{1}{x}=-\frac{2}{3}\\\\\frac{1}{x(2x+3)}=\frac{1}{3x}-\frac{2}{3(2x+3)}$