Integrals: Common Mistakes and Fixes (13)

Question

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

Solution — Step by Step

Final answer: $\boxed{\mathbf{\dfrac{1}{3} \tan^{-1}\left(\dfrac{x+2}{3}\right) + C}}$.

Why This Works

Whenever the denominator is a quadratic with negative discriminant (no real roots), completing the square turns it into the form $(u^2 + a^2)$, which has the inverse-tangent integral as its antiderivative.

If the discriminant were positive, partial fractions would be the move. If zero, completing the square gives $(x - r)^2$ and the integral becomes $-1/(x-r)$.

So the discriminant of the denominator tells you the technique:

• $\Delta < 0$: complete the square, $\tan^{-1}$.
• $\Delta = 0$: complete the square, perfect-square integration.
• $\Delta > 0$: partial fractions, log of factors.

Alternative Method

If we recognize the form immediately, we can shortcut: $\int dx/((x+2)^2 + 9)$ matches $\int du/(u^2 + 9)$ at sight, no substitution needed mentally. With practice, this becomes a one-liner.

Common Mistake