Integrals: Common Mistakes and Fixes (1)

Question

Evaluate $\displaystyle \int \frac{1}{x^2 + 4x + 13}\, dx$.

Solution — Step by Step

Final answer: $\frac{1}{3}\tan^{-1}\dfrac{x+2}{3} + C$.

Why This Works

Quadratic denominators with no real roots (discriminant negative) always reduce to the $u^2 + a^2$ form via completing the square. The arctangent integral is the standard answer.

If the discriminant were positive, the quadratic would factor and you would use partial fractions instead. If zero, you would get $1/(x+a)^2$ leading to $-1/(x+a)$.

Alternative Method

Look up the formula $\int \frac{dx}{ax^2 + bx + c} = \frac{2}{\sqrt{4ac - b^2}}\tan^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}}$ when $4ac > b^2$. Plug $a=1, b=4, c=13$: $4ac - b^2 = 52 - 16 = 36$, $\sqrt{36} = 6$. Answer: $\frac{2}{6}\tan^{-1}\frac{2x+4}{6} = \frac{1}{3}\tan^{-1}\frac{x+2}{3}$. Same result.