Integrals · Lesson 4 of 9

Standard Integrals Involving Quadratic Expressions

“Recognise six useful standard forms and transform general quadratic expressions into them.”

Learning Objectives

• Derive the six reciprocal quadratic and reciprocal square-root standard forms. • Choose a suitable trigonometric substitution and handle its interval restrictions. • Complete the square to transform a general quadratic. • Split a linear numerator into a denominator-derivative part and a remainder. • Distinguish logarithmic and inverse-trigonometric answers by the transformed form.

Recognising the shape beneath the coefficients

An expression such as $x^{2} - 6 x + 13$ may initially look unrelated to $1 + x^{2}$ . Completing the square reveals $(x - 3)^{2} + 4$ , so a shift and a scale reduce it to the inverse-tangent form. This lesson develops the standard forms first, then shows how to recognise them inside more general expressions.

Throughout the standard formulas take $a > 0$ . Logarithmic expressions include absolute values, and each formula is used on an interval where its integrand is real and defined. These restrictions are part of the formula, even when routine calculations do not repeat them on every line.

Reciprocals of quadratic expressions

The sign between the square terms determines the strategy. A difference of squares factors into linear terms, leading to logarithms. A sum of squares does not factor into real linear terms; a tangent substitution instead leads to an inverse tangent.

Three reciprocal quadratic formsLaTeX

\int \frac{d x}{x ^{2} - a ^{2}} \int \frac{d x}{a ^{2} - x ^{2}} \int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{2 a} ln \frac{x - a}{x + a} + C, = \frac{1}{2 a} ln \frac{a + x}{a - x} + C, = \frac{1}{a} tan^{- 1} (\frac{x}{a}) + C .

The first two exclude

x = \pm a

. The sum-of-squares integrand is defined for every real

x

For the first form, factor $x^{2} - a^{2} = (x - a) (x + a)$ and verify $1/ (x^{2} - a^{2}) = [1/ (x - a) - 1/ (x + a)] / (2 a)$ . Integrate the two logarithmic derivatives and combine their logarithms. For the second, use $1/ (a^{2} - x^{2}) = [1/ (a - x) + 1/ (a + x)] / (2 a)$ . The integral of $1/ (a - x)$ is $- ln ∣ a - x ∣$ , so the sign produces the displayed ratio.

For the third form put $x = a tan θ$ , taking $- π /2 < θ < π /2$ . Then $d x = a sec^{2} θ d θ$ and $x^{2} + a^{2} = a^{2} sec^{2} θ$ . The quotient reduces to $d θ / a$ , whose integral is $θ / a$ . Substituting $θ = tan^{- 1} (x / a)$ gives the formula.

Reciprocals of square roots

A square root changes which factor cancels after substitution. The identities $1 + tan^{2} θ = sec^{2} θ$ and $1 - sin^{2} θ = cos^{2} θ$ let us turn the square root into a single trigonometric factor. Choose an angle interval where the sign of that factor is known; $u^{2}$ means $∣ u ∣$ .

Three reciprocal square-root formsLaTeX

\int \frac{d x}{x ^{2} - a ^{2}} \int \frac{d x}{a ^{2} - x ^{2}} \int \frac{d x}{x ^{2} + a ^{2}} = ln x + x^{2} - a^{2} + C, = sin^{- 1} (\frac{x}{a}) + C, = ln x + x^{2} + a^{2} + C .

The first uses intervals with

∣ x ∣ > a

, the second

∣ x ∣ < a

, and the third allows all real

x

. Differentiation verifies the logarithmic formula on each permitted interval.

To derive the first on $x > a$ , put $x = a sec θ$ with $0 < θ < π /2$ . Then $d x = a sec θ tan θ d θ$ and the root is $a tan θ$ , leaving $\int sec θ d θ$ . Its logarithmic answer becomes $ln ∣ x + x^{2} - a^{2} ∣ - ln a + C$ ; absorb $- ln a$ into $C$ . The displayed expression also differentiates correctly on $x < - a$ .

For the second put $x = a sin θ$ with $- π /2 < θ < π /2$ . Both $d x$ and the root contain $a cos θ$ , so the integral becomes $\int d θ = θ + C$ . For the third put $x = a tan θ$ on the principal tangent interval: $d x = a sec^{2} θ d θ$ and the root is $a sec θ$ . Again the integral becomes a secant integral, producing the logarithmic expression after substitution back.

Example — Recognising two basic forms

Problem
Evaluate $\int d x / (x^{2} - 16)$ and $\int d x / 2 x - x^{2}$ .

1.In the first, $a = 4$ , so the result is $\frac{1}{8} ln ∣ (x - 4) / (x + 4) ∣ + C$ , with $x \neq = \pm 4$ .
2.In the second, complete the square: $2 x - x^{2} = 1 - (x - 1)^{2}$ . Put $t = x - 1$ and $d t = d x$ .
3.The result is $sin^{- 1} (x - 1) + C$ , valid for the integrand on $0 < x < 2$ .

Completing the square systematically

Completing the square is a change of viewpoint, not a new integration rule. It centres the quadratic at its turning point and leaves a constant whose sign identifies the standard form. A repeated-square case must be treated separately rather than forced into a formula involving division by a zero parameter.

Completing a general squareLaTeX

a x^{2} + b x + c = a [(x + \frac{b}{2 a})^{2} + \frac{4 a c - b ^{2}}{4 a ^{2}}] (a \neq = 0)

Set

t = x + b / (2 a)

, so

d t = d x

. For a reciprocal quadratic the factor outside is

1/ a

. For a square root, keep track of the sign of

a

and of the region where the radicand is positive.

If the remaining constant is positive, a sum of squares appears; if it is negative, a difference of squares appears. If it is zero, a reciprocal quadratic reduces to a constant multiple of $t^{- 2}$ , handled by the power rule. A reciprocal square root of a repeated square contains $1/∣ t ∣$ , so it is evaluated on one side of $t = 0$ at a time.

Example — A shifted sum of squares

Problem
Evaluate $\int d x / (x^{2} - 6 x + 13)$ .

1.Write the denominator as $(x - 3)^{2} + 4$ . Put $t = x - 3$ .
2.The integral becomes $\int d t / (t^{2} + 2^{2})$ .
3.Use the inverse-tangent form: $\frac{1}{2} tan^{- 1} ((x - 3) /2) + C$ .

Example — A shifted difference of squares

Problem
Evaluate $\int d x / (3 x^{2} + 13 x - 10)$ .

1.Write $3 x^{2} + 13 x - 10 = 3 [(x + 13/6)^{2} - (17/6)^{2}]$ .
2.With $t = x + 13/6$ , factor $1/3$ outside and use the difference-of-squares formula with $a = 17/6$ . The multiplying coefficient is $(1/3) / (2 \cdot 17/6) = 1/17$ .
3.Substitute back to obtain $\frac{1}{17} ln ∣ (3 x - 2) / (x + 5) ∣ + C$ . A constant factor inside a logarithm is absorbed into $C$ . The original denominator excludes $x = 2/3$ and $x = - 5$ .

Example — A quadratic under a square root

Problem
Evaluate $\int d x / 5 x^{2} - 2 x$ .

1.Complete the square: $5 x^{2} - 2 x = 5 [(x - 1/5)^{2} - (1/5)^{2}]$ . The root contributes a factor $5$ .
2.Put $t = x - 1/5$ . The integral is $(1/ 5) \int d t / t^{2} - (1/5)^{2}$ .
3.Therefore the result is $\frac{1}{5} ln ∣ x - 1/5 + x^{2} - 2 x /5 ∣ + C$ . Use it on intervals $x < 0$ or $x > 2/5$ , where the integrand is real and defined.

When the numerator is almost a derivative

A linear numerator can be split into two parts: one proportional to the derivative of the quadratic, and one constant remainder. The first part integrates immediately by substitution; the second is one of the standard forms just developed. Matching coefficients makes this split systematic.

Splitting the numeratorLaTeX

p x + q = A (2 a x + b) + B, A = \frac{p}{2 a}, B = q - \frac{p b}{2 a}

The letters

p, q, a, b, c

are fixed coefficients. The same split works over

a x^{2} + b x + c

and over its square root; the derivative part yields a logarithm in the first case and a square root in the second.

Example — A linear numerator over a quadratic

Problem
Evaluate $\int (x + 2) / (2 x^{2} + 6 x + 5) d x$ .

1.The denominator derivative is $4 x + 6$ . Write $x + 2 = \frac{1}{4} (4 x + 6) + \frac{1}{2}$ .
2.The integral is $\frac{1}{4} ln (2 x^{2} + 6 x + 5) + \frac{1}{2} \int d x / (2 x^{2} + 6 x + 5)$ . The quadratic is positive, so absolute values are optional here.
3.Complete the square: $2 x^{2} + 6 x + 5 = 2 [(x + 3/2)^{2} + (1/2)^{2}]$ . Its reciprocal integral is $tan^{- 1} (2 x + 3)$ .
4.Combine the parts: $\frac{1}{4} ln (2 x^{2} + 6 x + 5) + \frac{1}{2} tan^{- 1} (2 x + 3) + C$ .

Example — A linear numerator over a square root

Problem
Evaluate $\int (x + 3) / 5 - 4 x - x^{2} d x$ .

1.Let $Q = 5 - 4 x - x^{2}$ , so $Q' = - 4 - 2 x$ . Since $x + 3 = - Q'/2 + 1$ , split the integral into $- \frac{1}{2} \int Q'/ Q d x + \int d x / Q$ .
2.For the first part, set $u = Q$ . Its value is $- Q$ .
3.For the second, $Q = 9 - (x + 2)^{2}$ , giving $sin^{- 1} ((x + 2) /3)$ .
4.The answer is $- 5 - 4 x - x^{2} + sin^{- 1} ((x + 2) /3) + C$ , used on $- 5 < x < 1$ .

Common mistake

Do not confuse $1/ (x^{2} + a^{2})$ with $1/ x^{2} + a^{2}$ . The first gives an inverse tangent and includes $1/ a$ ; the second gives a logarithm. Also, $a^{2} u^{2} = ∣ a ∣∣ u ∣$ , so signs cannot be dropped without an interval argument.

Quick check

Which standard form produces $\frac{1}{a} tan^{- 1} (x / a) + C$ ?

Quick check

What is the completed-square form of $x^{2} + 4 x + 7$ ?

Quick check

What is $\int d x / 9 - x^{2}$ ?

Quick check

In $x + 2 = A (4 x + 6) + B$ , what are $A$ and $B$ ?

Quick check

On which intervals is $1/ x^{2} - 4$ real and defined?

Practice Problems

Find $\int 3 x^{2} / (x^{6} + 1) d x$ .
Find $\int d x / (9 - 25 x^{2})$ on an appropriate interval.
Find $\int d x / (x^{2} + 2 x + 2)$ .
Find $\int d x / 7 - 6 x - x^{2}$ .
Find $\int (5 x - 2) / (1 + 2 x + 3 x^{2}) d x$ .
Find $\int (x + 2) / 4 x - x^{2} d x$ .
Compare the substitutions and resulting answer types for $\int d x / (x^{2} + 4)$ and $\int d x / x^{2} + 4$ .
Find $\int d x / (x^{2} + 6 x + 9)$ and explain why the repeated-square case needs no inverse tangent.

Key Takeaways

Reciprocal quadratic forms lead to logarithms or inverse tangents. Reciprocal square-root forms lead to logarithms or inverse sines. Completing the square reveals the applicable form. A linear numerator can be separated into a quadratic-derivative part and a constant remainder. Scaling factors and domain restrictions are essential parts of every calculation.

Previous · Lesson 3

Integrating with Trigonometric Identities

Next · Lesson 5

Integration by Partial Fractions

Recognising the shape beneath the coefficients

Reciprocals of quadratic expressions

Reciprocals of square roots

Completing the square systematically

When the numerator is almost a derivative

Quiz

Practice Problems

Key Takeaways