Completing the square for a shifted quadratic, integrals from easy to advanced lesson

The denominator of $1/(x^2 + 4x + 8)$ has discriminant $16 - 32 = -16$ , so it has no real roots and does not factor into real linear pieces, and partial fractions with linear denominators do not apply. Completing the square writes it as $(x + 2)^2 + 4$ , a shifted and scaled copy of $1 + u^2$ , and $1/(1 + u^2)$ is exactly the derivative of the inverse tangent. This page explains completing the square, irreducible quadratics and the discriminant test, the inverse tangent antiderivative and the substitution $u = (x + 2)/2$ , and it evaluates the integral from $0$ to $2$ to $\tfrac12(\arctan 2 - \pi/4)$ .

Key terms

Completing the square

Definition For real numbers $a \ne 0$ , $b$ and $c$ and every real $x$ ,

\begin{aligned} &ax^2 + bx + c \\[0.4em] &\quad = a\Bigl(x + \frac{b}{2a}\Bigr)^2 \\[0.4em] &\qquad + c - \frac{b^2}{4a}. \end{aligned}

The identity follows by expanding the square, and it changes the form of the quadratic, not its value. With $h = -b/(2a)$ and $k = c - b^2/(4a)$ it reads $a(x - h)^2 + k$ , so the graph is a parabola with vertex $(h, k)$ . For $a = 1$ the rule is to add and subtract the square of half the coefficient of $x$ .

In plain words Half the coefficient of $x$ , squared, is the constant that completes a perfect square. Put it inside the square and take it back outside, and the expression keeps its value for every $x$ while its center becomes visible.

Example For the integral of this page, half of $4$ is $2$ , and $(x + 2)^2 = x^2 + 4x + 4$ , so $x^2 + 4x + 8 = (x + 2)^2 + 4$ . This is at least $4$ for every real $x$ , so the integrand is continuous on $[0, 2]$ . Factoring out $4$ gives $4\bigl(1 + ((x + 2)/2)^2\bigr)$ , and the substitution $u = (x + 2)/2$ has $dx = 2\,du$ and moves the bounds $0$ and $2$ to $1$ and $2$ . Hence

\begin{aligned} &\int_0^2 \frac{dx}{x^2 + 4x + 8} \\[0.4em] &\quad = \int_1^2 \frac{2\,du}{4(1 + u^2)} \\[0.4em] &\quad = \frac12\Bigl[\arctan u\Bigr]_1^2 \\[0.4em] &\quad = \frac12\Bigl(\arctan 2 - \frac{\pi}{4}\Bigr), \end{aligned}

with the inverse tangent in radians and taking values in $(-\pi/2, \pi/2)$ . The value is about $0.1609$ . In the original variable the antiderivative is $F(x) = \tfrac12\arctan\bigl((x + 2)/2\bigr)$ , and the chain rule confirms it,

\begin{aligned} F'(x) &= \frac{1/4}{1 + (x + 2)^2/4} \\[0.4em] &= \frac{1}{(x + 2)^2 + 4}. \end{aligned}

As a rough check, the integrand decreases on $[0, 2]$ from $1/8$ to $1/20$ , so over this interval of length $2$ the integral lies between $1/10$ and $1/4$ .

Irreducible quadratic

Definition A quadratic $ax^2 + bx + c$ with real coefficients and $a \ne 0$ is irreducible over the real numbers if it cannot be written as a product of two linear factors with real coefficients. This happens exactly when it has no real root, which is the case exactly when $b^2 - 4ac < 0$ . Over the complex numbers every quadratic factors, and an irreducible real quadratic has two complex conjugate roots. In a partial fraction decomposition, an irreducible quadratic factor gets a numerator of the form $Bx + C$ .

In plain words An irreducible quadratic never meets the $x$ -axis, so it cannot be split into real linear factors. Its reciprocal integrates to an inverse tangent, and a linear numerator adds a logarithm of the quadratic.

Example The quadratic $x^2 + 4x + 8$ is irreducible. Its roots are the complex numbers $-2 \pm 2i$ , and $x^2 + 4x + 8 = (x + 2 - 2i)(x + 2 + 2i)$ . By contrast $x^2 + 4x + 3 = (x + 1)(x + 3)$ is reducible. For a linear numerator, split off a multiple of the derivative $2x + 4$ of the quadratic,

\begin{aligned} &\frac{x + 4}{x^2 + 4x + 8} \\[0.4em] &\quad = \frac{\tfrac12(2x + 4)}{x^2 + 4x + 8} \\[0.4em] &\qquad + \frac{2}{x^2 + 4x + 8}. \end{aligned}

The first part integrates to $\tfrac12\ln(x^2 + 4x + 8)$ , with no absolute value because the quadratic is positive, and the second to $\arctan\bigl((x + 2)/2\bigr)$ . So the antiderivatives on the whole real line are $\tfrac12\ln(x^2 + 4x + 8) + \arctan\bigl((x + 2)/2\bigr) + C$ .

Negative discriminant

Definition The discriminant of $ax^2 + bx + c$ , with $a \ne 0$ , is $\Delta = b^2 - 4ac$ . Completing the square gives

\begin{aligned} &ax^2 + bx + c \\[0.4em] &\quad = a\Bigl(x + \frac{b}{2a}\Bigr)^2 - \frac{\Delta}{4a}. \end{aligned}

If $\Delta < 0$ , the second term $-\Delta/(4a)$ is not $0$ and has the sign of $a$ , and the first term is $0$ or has the sign of $a$ . So the quadratic is never $0$ and has the sign of $a$ for every real $x$ . Its value $-\Delta/(4a)$ at $x = -b/(2a)$ is the minimum when $a > 0$ and the maximum when $a < 0$ . If $\Delta = 0$ , the quadratic is $a\bigl(x + b/(2a)\bigr)^2$ , with one repeated real root, and if $\Delta > 0$ , it has the two distinct real roots $\bigl(-b \pm \sqrt{\Delta}\bigr)/(2a)$ .

In plain words A negative discriminant means that the quadratic has no real roots and keeps one sign. Its reciprocal is then continuous on the whole real line, so an integral of the reciprocal over any interval has no singularity.

Example For $x^2 + 4x + 8$ , the discriminant is $16 - 32 = -16$ . The minimum value is $16/4 = 4$ , at $x = -2$ , where the integrand $1/(x^2 + 4x + 8)$ has its highest point $1/4$ . The interval $[0, 2]$ lies to the right of this peak, where the integrand decreases. For $2x^2 - 4x + 5$ , the discriminant is $16 - 40 = -24$ , and the minimum value is $24/8 = 3$ , at $x = 1$ . For $x^2 + 4x + 3$ , the discriminant is $16 - 12 = 4 > 0$ , and the roots are $-1$ and $-3$ .

Inverse tangent antiderivative

Definition For every real $u$ , the inverse tangent $\arctan u$ , also written $\tan^{-1} u$ , is the unique angle $\theta$ in the open interval $(-\pi/2, \pi/2)$ with $\tan\theta = u$ , measured in radians. It is differentiable on the whole real line, with

\frac{d}{du}\arctan u = \frac{1}{1 + u^2},

because $u = \tan\theta$ gives $du/d\theta = \sec^2\theta = 1 + \tan^2\theta = 1 + u^2$ , which is positive, and the derivative of an inverse function is the reciprocal of this rate. Consequently, for a constant $a > 0$ and on the whole real line,

\begin{aligned} &\int \frac{du}{u^2 + a^2} \\[0.4em] &\quad = \frac1a\arctan\frac{u}{a} + C. \end{aligned}

In plain words The function $1/(1 + u^2)$ is the rate of change of an angle. Its integral from $p$ to $q$ is the change in the angle whose tangent is $u$ , from $\arctan p$ to $\arctan q$ .

Example Since $\arctan 1 = \pi/4$ ,

\begin{aligned} &\int_1^2 \frac{du}{1 + u^2} = \arctan 2 - \frac{\pi}{4} \\[0.4em] &\quad \approx 1.1071 - 0.7854 \\[0.4em] &\quad \approx 0.3218. \end{aligned}

The value $\arctan 2$ is not a standard angle, so it stays in the exact answer. The angles must be in radians. In degrees the same difference would be about $18.43$ , which is not the integral.

Shift and scale substitution

Definition For constants $h$ and $k > 0$ , the substitution $u = (x - h)/k$ , with $x = h + ku$ and $dx = k\,du$ , turns $(x - h)^2 + k^2$ into $k^2(1 + u^2)$ . A definite integral from $x = p$ to $x = q$ becomes an integral from $u = (p - h)/k$ to $u = (q - h)/k$ , and on the whole real line

\begin{aligned} &\int \frac{dx}{(x - h)^2 + k^2} \\[0.4em] &\quad = \frac1k\arctan\frac{x - h}{k} + C. \end{aligned}

In plain words The shift moves the center of the quadratic to $u = 0$ , and dividing by $k$ makes the constant term $1$ . The width factor $dx = k\,du$ and the height factor $1/k^2$ combine into the single factor $1/k$ in front. Forgetting the width factor is the most common slip.

Example Completing the square gives $x^2 - 2x + 10 = (x - 1)^2 + 9$ , so $h = 1$ and $k = 3$ . With $u = (x - 1)/3$ , the bounds $1$ and $4$ become $0$ and $1$ , and

\begin{aligned} &\int_1^4 \frac{dx}{x^2 - 2x + 10} \\[0.4em] &\quad = \frac13\int_0^1 \frac{du}{1 + u^2} \\[0.4em] &\quad = \frac13 \cdot \frac{\pi}{4} = \frac{\pi}{12}, \end{aligned}

which is about $0.2618$ . Leaving out $dx = 3\,du$ would give $\pi/36$ , three times too small. As a check, the integrand decreases on $[1, 4]$ from $1/9$ to $1/18$ , so over this interval of length $3$ the integral lies between $1/6$ and $1/3$ . The integral from $0$ to $2$ above is the case $h = -2$ and $k = 2$ .

Check yourself

Evaluate the integral below.

\int_{-1}^{1} \frac{dx}{x^2 + 2x + 5}

Show the answer

Completing the square gives $x^2 + 2x + 5 = (x + 1)^2 + 4$ , which is at least $4$ , so the integrand is continuous everywhere. With $u = (x + 1)/2$ and $dx = 2\,du$ , the bounds $-1$ and $1$ become $0$ and $1$ , and

\begin{aligned} &\int_{-1}^{1} \frac{dx}{x^2 + 2x + 5} \\[0.4em] &\quad = \frac{2}{4}\int_0^1 \frac{du}{1 + u^2} \\[0.4em] &\quad = \frac12 \cdot \frac{\pi}{4} = \frac{\pi}{8}, \end{aligned}

which is about $0.3927$ . The integrand decreases from $1/4$ to $1/8$ on this interval of length $2$ , so the value lies between $1/4$ and $1/2$ , and it does.

Evaluate the integral below, taking care with the leading coefficient.

\int_0^1 \frac{dx}{4x^2 + 4x + 2}

Show the answer

The discriminant is $16 - 32 = -16 < 0$ , so the quadratic is irreducible. Completing the square with the leading coefficient factored out gives $4x^2 + 4x + 2 = 4\bigl(x + \tfrac12\bigr)^2 + 1 = (2x + 1)^2 + 1$ . With $u = 2x + 1$ and $dx = \tfrac12\,du$ , the bounds $0$ and $1$ become $1$ and $3$ , so

\begin{aligned} &\int_0^1 \frac{dx}{4x^2 + 4x + 2} \\[0.4em] &\quad = \frac12\int_1^3 \frac{du}{1 + u^2} \\[0.4em] &\quad = \frac12\Bigl(\arctan 3 - \frac{\pi}{4}\Bigr), \end{aligned}

which is about $0.2318$ . The integrand decreases from $1/2$ to $1/10$ on an interval of length $1$ , consistent with the value. Dropping the factor $\tfrac12$ that comes from $dx$ would give about $0.4636$ , twice the correct value.

Someone completes the square in $x^2 + 4x + 3$ and writes $\arctan(x + 2) + C$ as the antiderivative of $1/(x^2 + 4x + 3)$ . What went wrong? Evaluate the integral of $1/(x^2 + 4x + 3)$ from $0$ to $1$ .

Show the answer

The discriminant is $16 - 12 = 4 > 0$ , so the quadratic is reducible, and completing the square gives $(x + 2)^2 - 1$ , a difference rather than a sum. The inverse tangent form needs $(x - h)^2 + k^2$ with $k > 0$ . Differentiating the claimed answer exposes the error, since the derivative of $\arctan(x + 2)$ is $1/\bigl((x + 2)^2 + 1\bigr) = 1/(x^2 + 4x + 5)$ . Partial fractions apply instead. From $x^2 + 4x + 3 = (x + 1)(x + 3)$ and $1 = A(x + 3) + B(x + 1)$ , the value $x = -1$ gives $A = 1/2$ and the value $x = -3$ gives $B = -1/2$ . The poles $-1$ and $-3$ lie outside $[0, 1]$ , where both factors are positive, so

\begin{aligned} &\int_0^1 \frac{dx}{x^2 + 4x + 3} \\[0.4em] &\quad = \frac12\Bigl[\ln\frac{x + 1}{x + 3}\Bigr]_0^1 \\[0.4em] &\quad = \frac12\Bigl(\ln\frac12 - \ln\frac13\Bigr) \\[0.4em] &\quad = \frac12\ln\frac32. \end{aligned}

After the shift $v = x + 2$ , this is the integral of $1/(v^2 - 1)$ from $2$ to $3$ , worked on the page on partial fractions.

Completing the square for a shifted quadratic

Key terms

Completing the square

Irreducible quadratic

Negative discriminant

Inverse tangent antiderivative

Shift and scale substitution

Check yourself

How this lesson connects

Builds on