Evaluate the integral: I = integral dx/(x²+3x)

Evaluate the integral: $$ I = \int \frac{dx}{x^2+3x}. $$

Step 1 · Factor and set up partial fractions

$$ x^2+3x = x(x+3), \qquad \frac{1}{x(x+3)} = \frac Ax + \frac{B}{x+3}. $$

Multiplying by $x(x+3)$: $1=A(x+3)+Bx$. Setting $x=0$ gives $A=\tfrac13$; setting $x=-3$ gives $B=-\tfrac13$.

Step 2 · Integrate

$$ I = \int \left(\frac{1}{3x} - \frac{1}{3(x+3)}\right) dx = \frac13\ln|x| - \frac13\ln|x+3| + C. $$

Step 3 · Combine logarithms

$$ I = \frac13\ln\left|\frac{x}{x+3}\right| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=0$ and $x=-3$, where the original integrand is undefined.
• This follows the identical pattern to Problem 187, with the second root shifted from $-2$ to $-3$; in general, $\int dx/[x(x+k)] = \tfrac1k\ln|x/(x+k)|+C$.
• The result can be checked by differentiating $\tfrac13\ln|x|-\tfrac13\ln|x+3|$: this gives $\tfrac{1}{3x}-\tfrac{1}{3(x+3)}=\tfrac{(x+3)-x}{3x(x+3)}=\tfrac{1}{x(x+3)}$, matching the integrand.