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

01 / LEARN

Understand the idea.

Why the power rule stops at minus one

The usual power rule fails for $x^{- 1}$ because its new exponent would be zero. Instead, use $\frac{d}{d x} ln ∣ x ∣ = \frac{1}{x}$ for $x \neq 0$ . The absolute value permits a real logarithm on either side of zero.

For a linear denominator $a x + b$ , divide the numerator coefficient by $a$ . The result holds on an interval avoiding $x = - \frac{b}{a}$ , with $a \neq 0$ . Constants may be chosen independently on the two separate intervals.

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

Simplify before choosing a rule

Simplify the integrand before integrating. Dividing each numerator term by a common denominator may reveal both an ordinary power and a reciprocal term. These need different antiderivative rules, even though they began in one fraction.

Check the denominator throughout a proposed definite-integral interval. An antiderivative formula on each side of a singularity does not justify subtracting endpoint values across it. The usual continuous-integrand theorem does not apply across an undefined point.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

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

Worked solution

1
Differentiate the logarithmic candidate.
$\frac{d}{d x} ln ∣ 2 x + 3 ∣ = \frac{2}{2 x + 3}$
The chain rule supplies numerator 2 rather than the required 4.
2
Scale the candidate and add the constant.
$\int \frac{4}{2 x + 3} d x = 2 ln ∣ 2 x + 3 ∣ + C$
Multiplying by 2 produces the required numerator.
3
State the permitted intervals.
$x < - \frac{3}{2} or x > - \frac{3}{2}$
The denominator cannot be zero; the formula is used on one such interval at a time.
4
Check the scaled derivative.
$\frac{d}{d x} (2 ln ∣ 2 x + 3 ∣ + C) = \frac{4}{2 x + 3}$
The derivative matches the original reciprocal expression wherever it is defined.

Answer

$2 ln ∣ 2 x + 3 ∣ + C$

EXAMPLE 02

Worked example

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

Worked solution

1
Divide each numerator term by the denominator.
$\frac{6 x^{2} + 1}{x} = 6 x + \frac{1}{x}$
The common denominator can be applied to each term of the sum because $x > 0$ .
2
Integrate the two different patterns.
$\begin{aligned} \int 6 x d x & = 3 x^{2} + C_{1} \\ \int \frac{1}{x} d x & = ln x + C_{2} \end{aligned}$
The first term uses the power rule; the positive domain allows the second logarithm without absolute values.
3
Combine and check.
$\begin{aligned} F (x) & = 3 x^{2} + ln x + C \\ F^{'} (x) & = 6 x + \frac{1}{x} \end{aligned}$
Linearity and the derivative check establish the required antiderivative on the stated domain.

Answer

$3 x^{2} + ln x + C$

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$ln ∣ 3 x - 1 ∣ + C$

The denominator derivative is 3, matching the numerator. Thus the antiderivative is $ln ∣ 3 x - 1 ∣ + C$ , valid on an interval avoiding $x = \frac{1}{3}$ .

Keep practising

6 worksheet questions · 18 marks

Open worksheet 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 $\int (x + \frac{2}{x}) d x$ on $x > 0$ .
3 marks
Find $\int \frac{x + 3}{x} d x$ on $x < 0$ .
3 marks
Can $\int_{- 1}^{1} \frac{1}{x} d x$ 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 $\int \frac{1}{x} d x = ln ∣ x ∣ + C$ .
$5 ln ∣ x ∣ + C$
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 $\int (x + \frac{2}{x}) d x$ on $x > 0$ .
3 marks
Use the power rule for $x$ and log rule for $\frac{1}{x}$ .
$\frac{x^{2}}{2} + 2 ln x + C$
Find $\int \frac{x + 3}{x} d x$ on $x < 0$ .
3 marks
Simplify to $1 + \frac{3}{x}$ .
The absolute value gives a real logarithm on $x < 0$ .
$x + 3 ln ∣ x ∣ + C$
Can $\int_{- 1}^{1} \frac{1}{x} d x$ 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 ↗

In this lesson

Understand the idea.

Why the power rule stops at minus one

Simplify before choosing a rule

See the method in action.

Worked example

Differentiate the logarithmic candidate.

Scale the candidate and add the constant.

State the permitted intervals.

Check the scaled derivative.

Worked example

Divide each numerator term by the denominator.

Integrate the two different patterns.

Combine and check.

Try it yourself.

Check your understanding

Keep practising