Answer
$$\int\frac{dx}{x^2+2x}=\frac{1}{2}\ln|x|-\frac{1}{2}\ln|x+2|+C$$
Work Step by Step
$$A=\int\frac{dx}{x^2+2x}$$
1) Express the integrand as a sum of partial fractions: $$\frac{1}{x^2+2x}=\frac{1}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}$$
Clear fractions: $$1=A(x+2)+Bx$$ $$1=Ax+Bx+2A$$ $$1=(A+B)x+2A$$
Equating coefficients of corresponding powers of $x$, we get:
$A+B=0$
$2A=1$
Calculation gives us $A=1/2$ and $B=-1/2$. Therefore, $$\frac{1}{x^2+2x}=\frac{1}{2x}-\frac{1}{2(x+2)}$$
2) Evaluate the integral: $$A=\frac{1}{2}\int\frac{dx}{x}-\frac{1}{2}\int\frac{dx}{x+2}$$ $$A=\frac{1}{2}\ln|x|-\frac{1}{2}\ln|x+2|+C$$
