Integrating with Long Division and Completing the Square

Quick answer

Some integrands match no rule until they are rewritten. When a rational function's numerator has degree at least that of its denominator, long division splits it into a polynomial plus a proper fraction: (x² + 1)/(x + 1) = x − 1 + 2/(x + 1). When a quadratic denominator has no real roots, completing the square turns it into u² + a², and ∫ du/(u² + a²) = (1/a) arctan(u/a) + C. Both steps are algebra done before the calculus starts.

What you'll learn

Use long division to integrate a top-heavy rational function
Complete the square in a quadratic denominator
Integrate 1/(u² + a²) with the arctangent
Split a linear numerator into a logarithm part and an arctangent part

Rewrite first, then integrate

The integral $\int \tfrac{x^2 + 1}{x + 1}\,dx$ matches no rule: the numerator is not the derivative of the denominator, and there is no quotient rule for integrals. The way in is algebra. Rewrite the integrand as a sum of pieces that each match a rule, then integrate the pieces.

Two rewrites handle most fractions of polynomials: long division and completing the square.

Long division

When the numerator’s degree is at least the denominator’s, divide:

\frac{x^2 + 1}{x + 1} = x - 1 + \frac{2}{x + 1}

Check by multiplying back: $(x + 1)(x - 1) + 2 = x^2 + 1$ . Each piece now integrates on sight.

Why dividing is the right move

A fraction whose numerator is at least as big as its denominator is like the improper fraction $\tfrac{7}{3}$ , which is $2 + \tfrac{1}{3}$ . Division splits off the whole part, a polynomial, which integrates term by term. What remains has a numerator of lower degree than its denominator, and that is the shape the log rule, substitution and the arctangent need. Divide until the numerator’s degree is less than the denominator’s; then integrate.

Completing the square

A quadratic denominator with no real roots, such as $x^2 + 4x + 5$ , cannot be factored. Complete the square instead: $x^2 + 4x + 5 = (x + 2)^2 + 1$ . With $u = x + 2$ , the integral becomes one the arctangent answers:

\int \frac{du}{u^2 + 1} = \arctan u + C \qquad\qquad \int \frac{du}{u^2 + a^2} = \frac{1}{a}\arctan\frac{u}{a} + C

The second form follows from the first with $u = aw$ .

y = 1/(x² + 1)
y = 1/((x + 2)² + 1)

Completing the square is a shift

The graph shows why this works: $\tfrac{1}{(x + 2)^2 + 1}$ is the curve $\tfrac{1}{x^2 + 1}$ moved $2$ units left, so its antiderivative is $\arctan x$ moved $2$ units left, $\arctan(x + 2)$ .

Worked examples

Common mistakes

Practice problems

Find $\displaystyle\int \frac{x^2 + 3x + 2}{x + 1}\,dx$ .

Answer

$\tfrac{x^2}{2} + 2x + C$

Full solution

The numerator factors: $x^2 + 3x + 2 = (x + 1)(x + 2)$ , so the fraction is $x + 2$ (for $x \ne -1$ ).
Find $\displaystyle\int \frac{x}{x + 2}\,dx$ .

Answer

$x - 2\ln\lvert x + 2 \rvert + C$

Full solution

$\tfrac{x}{x + 2} = \tfrac{(x + 2) - 2}{x + 2} = 1 - \tfrac{2}{x + 2}$ .
Find $\displaystyle\int \frac{x^3 - 1}{x - 1}\,dx$ .

Answer

$\tfrac{x^3}{3} + \tfrac{x^2}{2} + x + C$

Full solution

Dividing gives $x^3 - 1 = (x - 1)(x^2 + x + 1)$ , so the integrand is $x^2 + x + 1$ .
Find $\displaystyle\int \frac{2x^2 + 1}{x^2 + 1}\,dx$ .

Answer

$2x - \arctan x + C$

Full solution

$\tfrac{2x^2 + 1}{x^2 + 1} = \tfrac{2(x^2 + 1) - 1}{x^2 + 1} = 2 - \tfrac{1}{x^2 + 1}$ .
Find $\displaystyle\int \frac{dx}{x^2 + 2x + 2}$ .

Answer

$\arctan(x + 1) + C$

Full solution

$x^2 + 2x + 2 = (x + 1)^2 + 1$ .
Find $\displaystyle\int \frac{dx}{x^2 + 9}$ .

Answer

$\tfrac{1}{3}\arctan\tfrac{x}{3} + C$

Full solution

Use $\int \tfrac{du}{u^2 + a^2} = \tfrac{1}{a}\arctan\tfrac{u}{a} + C$ with $a = 3$ .
Find $\displaystyle\int \frac{dx}{x^2 - 4x + 8}$ .

Answer

$\tfrac{1}{2}\arctan\tfrac{x - 2}{2} + C$

Full solution

$x^2 - 4x + 8 = (x - 2)^2 + 4$ , so $u = x - 2$ and $a = 2$ .
Evaluate $\displaystyle\int_0^1 \frac{dx}{x^2 + 1}$ .

Answer

$\tfrac{\pi}{4}$

Full solution

$\arctan 1 - \arctan 0 = \tfrac{\pi}{4} - 0$ .
Find $\displaystyle\int \frac{2x + 2}{x^2 + 2x + 5}\,dx$ .

Answer

$\ln\left(x^2 + 2x + 5\right) + C$

Full solution

The numerator is the derivative of the denominator, so substitute $u = x^2 + 2x + 5$ . No arctangent is needed. The absolute value can go, since $x^2 + 2x + 5 = (x + 1)^2 + 4 > 0$ .
A student writes $\displaystyle\int \frac{dx}{x^2 + 6x + 10} = \ln\left\lvert x^2 + 6x + 10 \right\rvert + C$ . What went wrong?

Hint

Differentiate the student’s answer. Does it give back the integrand?

Answer

The numerator is not the derivative of the denominator. Completing the square gives $\arctan(x + 3) + C$ .

Full solution

The derivative of the student’s answer is $\tfrac{2x + 6}{x^2 + 6x + 10}$ , not $\tfrac{1}{x^2 + 6x + 10}$ .

Instead, $x^2 + 6x + 10 = (x + 3)^2 + 1$ , and $\int \tfrac{dx}{(x + 3)^2 + 1} = \arctan(x + 3) + C$ .

Frequently asked questions

When should I use long division before integrating?

When the numerator's degree is at least the denominator's. Divide first, so that what is left is a polynomial plus a fraction with a smaller numerator.

What is the integral of 1/(x² + a²)?

(1/a) arctan(x/a) + C.

When do I complete the square?

When the denominator is a quadratic with no real roots, such as x² + 4x + 5. Completing the square makes it (x + 2)² + 1, which leads to the arctangent.

What if the numerator is x + 3 instead of 1?

Split it into a multiple of the denominator's derivative, which integrates to a logarithm, plus a constant, which integrates to an arctangent.

How do I know whether to use a logarithm or an arctangent?

If the numerator is a constant multiple of the denominator's derivative, substitute and get a logarithm. If it is a constant over an irreducible quadratic, complete the square and get an arctangent.

What to learn next

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

Rewrite first, then integrate

Long division

Why dividing is the right move

Completing the square

Worked examples

Common mistakes

Practice problems

Frequently asked questions

When should I use long division before integrating?

What is the integral of 1/(x² + a²)?

When do I complete the square?

What if the numerator is x + 3 instead of 1?

How do I know whether to use a logarithm or an arctangent?

What to learn next

More Calculus lessons