Evaluate ∫dx/(x² + 4x + 13) using completing the square

Question

Evaluate:

\int \frac{dx}{x^2 + 4x + 13}

(JEE Main 2021, similar pattern)

Solution — Step by Step

x^2 + 4x + 13 = (x^2 + 4x + 4) + 9 = (x + 2)^2 + 3^2

The denominator is now in the standard form $u^2 + a^2$ .

Let $u = x + 2$ , so $du = dx$ .

\int \frac{dx}{(x+2)^2 + 9} = \int \frac{du}{u^2 + 3^2}

\int \frac{du}{u^2 + a^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C

= \frac{1}{3}\tan^{-1}\left(\frac{u}{3}\right) + C

Substitute back $u = x + 2$ :

\boxed{\int \frac{dx}{x^2 + 4x + 13} = \frac{1}{3}\tan^{-1}\left(\frac{x+2}{3}\right) + C}

Why This Works

The trick is recognising that any quadratic $ax^2 + bx + c$ (with $a > 0$ and discriminant $< 0$ ) can be written as $(x + p)^2 + q^2$ by completing the square. Once in this form, we match it to the standard $\int \frac{du}{u^2 + a^2}$ result, which comes from the derivative of $\tan^{-1}$ .

The condition “discriminant $< 0$ ” is important — it means the quadratic has no real roots, so the denominator never becomes zero. Here, $\Delta = 16 - 52 = -36 < 0$ , confirming the integral is well-defined for all real $x$ .

Alternative Method — Direct application without substitution

You can skip the explicit substitution and write:

\int \frac{dx}{(x+2)^2 + 3^2} = \frac{1}{3}\tan^{-1}\left(\frac{x+2}{3}\right) + C

directly, treating $x + 2$ as the variable. This saves a line of work but requires you to be comfortable with the formula.

Completing the square is the gateway step for all integrals of the form $\int \frac{dx}{ax^2 + bx + c}$ . Depending on the sign of the discriminant, you’ll end up with either $\tan^{-1}$ (discriminant < 0) or $\log$ (discriminant > 0). Master this one technique and you can handle an entire family of integrals.

Common Mistake

A frequent error: writing $(x+2)^2 + 9$ but then using $a = 9$ instead of $a = 3$ in the formula. Remember, the form is $u^2 + a^2$ , not $u^2 + a$ . Here $a^2 = 9$ , so $a = 3$ . Using $a = 9$ gives $\frac{1}{9}\tan^{-1}\left(\frac{x+2}{9}\right)$ — wrong by a factor of 3 in both the coefficient and the argument.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct application without substitution

Common Mistake

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