Antiderivatives involving reciprocal linear expressions, linked back to logarithmic derivatives | Year 12

01 / LEARN

The idea, step by step

Why the power rule stops at minus one

Trying the power rule on $x^{- 1}$ would divide by zero. Instead use the known derivative of $ln ∣ x ∣$ , which is $1 / x$ wherever $x \neq 0$ . The antiderivative is valid on an interval that does not cross zero.

For a linear denominator, differentiating the logarithm introduces the coefficient of $x$ . Divide by that coefficient. An absolute value allows either positive or negative denominator on a suitable interval.

\int \frac{k}{a x + b} d x = \frac{k}{a} ln ∣ a x + b ∣ + C (a \neq 0)

Simplify before choosing a rule

A numerator may let you split a fraction into simpler terms. For instance, $(x^{2} + 1) / x = x + 1 / x$ . Integrate the power term and logarithmic term with their different rules.

An antiderivative formula is not permission to evaluate across a point where the integrand is undefined. At this level, definite integrals must use an interval on which the supplied integrand is continuous. Check the denominator first.

A common trap

Treating $1 / x$ as if its integral were $x^{0} / 0$ , or writing $ln (a x + b)$ on a negative-denominator interval.

Instead: Use the logarithmic rule and state the allowed interval; $ln ∣ a x + b ∣$ handles either sign away from the zero.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Find $\int \frac{4}{2 x + 3} d x$ .

Use $ln ∣ 2 x + 3 ∣$ , whose derivative is $2 / (2 x + 3)$ .
Multiply by 2 to obtain numerator 4.
The formula applies on either side of $x = - 3 / 2$ , not across it.

2 ln ∣ 2 x + 3 ∣ + C

Example 2

Worked example

Find $\int \frac{6 x^{2} + 1}{x} d x$ for $x > 0$ .

Split the integrand into $6 x + 1 / x$ .
Integrate each term using its own rule.
Since $x > 0$ , $ln x$ is valid without absolute-value notation.

3 x^{2} + ln x + C

Check your understanding

Find $\int \frac{3}{3 x - 1} d x$ .

Show answer

$ln ∣ 3 x - 1 ∣ + C$

The numerator matches the derivative of the linear denominator.

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find $\int \frac{5}{x} d x$ .
3 marks
Find $\int \frac{2}{4 x + 1} d x$ .
3 marks
Find $\int \frac{1}{5 - 2 x} d x$ .
3 marks
Find $\int (x + 2 / x) d x$ on $x > 0$ .
3 marks
Find $\int \frac{x + 3}{x} d x$ on $x < 0$ .
3 marks
Can $\int_{- 1}^{1} 1 / x d x$ be evaluated by substituting into $ln ∣ x ∣$ under the ordinary definite-integral rules? Explain.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $\int \frac{5}{x} d x$ .
3 marks
Use $\int 1 / x d x = ln ∣ x ∣ + C$ .
$5 ln ∣ x ∣ + C$
Find $\int \frac{2}{4 x + 1} d x$ .
3 marks
Divide 2 by the input coefficient 4.
$\frac{1}{2} ln ∣ 4 x + 1 ∣ + C$
Find $\int \frac{1}{5 - 2 x} d x$ .
3 marks
The denominator derivative is $- 2$ ; divide by $- 2$ .
$- \frac{1}{2} ln ∣ 5 - 2 x ∣ + C$
Find $\int (x + 2 / x) d x$ on $x > 0$ .
3 marks
Use the power rule for $x$ and log rule for $1 / x$ .
$x^{2} / 2 + 2 ln x + C$
Find $\int \frac{x + 3}{x} d x$ on $x < 0$ .
3 marks
Simplify to $1 + 3 / x$ .
The absolute value gives a real logarithm on $x < 0$ .
$x + 3 ln ∣ x ∣ + C$
Can $\int_{- 1}^{1} 1 / x d x$ be evaluated by substituting into $ln ∣ x ∣$ under the ordinary definite-integral rules? Explain.
3 marks
The integrand is undefined at 0 inside the interval.
The continuity condition for the usual fundamental-theorem evaluation fails.
No.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods integration and logarithmic antiderivatives; original learning and assessment material, with by-hand methods and explicitly described numerical models.

Official curriculum source ↗

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