Integration by Completing the Square Calculator

Integrating by Completing the Square and Substitution

In calculus, integrating rational functions or radical expressions containing a quadratic denominator ($ax^2 + bx + c$) often cannot be done using basic power rules alone. When the quadratic denominator does not factor neatly into simple linear terms or when it is inside a square root, completing the square transforms the quadratic expression into vertex form:

$$ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)$$

By substituting $u = x + \frac{b}{2a}$ (where $du = dx$), the integral reduces directly into well-known standard forms involving inverse trigonometric functions like $\arctan(u)$, natural logarithms $\ln|u|$, or inverse hyperbolic functions.

Step-by-Step Integration Method

Step 1: Complete the Square on the Quadratic Expression

Given the quadratic polynomial $ax^2 + bx + c$:

1. Factor out $a$ from the first two terms: $a\left(x^2 + \frac{b}{a}x\right) + c$.
2. Add and subtract $\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$ inside the parentheses.
3. Express as a perfect square plus a remainder: $a(x - h)^2 + k$, where $h = -\frac{b}{2a}$ and $k = c - \frac{b^2}{4a}$.

You can also use our standalone Completing the Square Calculator to practice the algebraic reduction independently.

Step 2: Split the Linear Numerator ($px + q$)

If the numerator contains an $x$-term ($p \neq 0$), decompose $px + q$ into a multiple of the denominator's derivative $(2ax + b)$ and an auxiliary constant:

$$px + q = \frac{p}{2a}(2ax + b) + \left(q - \frac{pb}{2a}\right)$$

The first part integrates by standard logarithmic substitution $\int \frac{f'(x)}{f(x)} dx = \ln|f(x)|$, while the second constant part is integrated using the completed square form.

Step 3: Apply the Standard Integration Formulas

Depending on the discriminant $\Delta = b^2 - 4ac$ and whether the denominator is rational or under a radical:

• Rational form with positive remainder ($k/a = M^2 > 0$): $$\int \frac{1}{u^2 + M^2} du = \frac{1}{M} \arctan\left(\frac{u}{M}\right) + C$$
• Rational form with negative remainder ($k/a = -M^2 < 0$): $$\int \frac{1}{u^2 - M^2} du = \frac{1}{2M} \ln\left|\frac{u - M}{u + M}\right| + C$$
• Radical form with $a > 0$: $$\int \frac{1}{\sqrt{u^2 + M^2}} du = \operatorname{arsinh}\left(\frac{u}{M}\right) + C = \ln\left|u + \sqrt{u^2 + M^2}\right| + C$$
• Radical form with $a < 0$: $$\int \frac{1}{\sqrt{M^2 - u^2}} du = \arcsin\left(\frac{u}{M}\right) + C$$

Worked Example

Evaluate the indefinite integral $\int \frac{1}{x^2 + 4x + 13} dx$:

1. Identify $a = 1, b = 4, c = 13$.
2. Complete the square: $x^2 + 4x + 13 = (x + 2)^2 - 4 + 13 = (x + 2)^2 + 9 = (x + 2)^2 + 3^2$.
3. Set $u = x + 2$, with $du = dx$ and $M = 3$.
4. Integrate: $\int \frac{1}{u^2 + 3^2} du = \frac{1}{3} \arctan\left(\frac{u}{3}\right) + C$.
5. Substitute back $u$: $\frac{1}{3} \arctan\left(\frac{x + 2}{3}\right) + C$.

Explore related inverse trigonometric evaluations with our Arctan Calculator and Arcsin Calculator.