Difficulty: Intermediate | Prerequisites: Basic integration techniques, u-substitution (Unit 4)
This is the first half of Calc AB Unit 5, covering the natural logarithm and how it appears in integration. You already know the derivative of ln x from earlier work; now you're going the other direction, recognising when an integral produces a natural log. This matters because a huge number of integrals in calculus (and in applied maths, physics, and engineering) reduce to ln once you spot the right structure. If you're comfortable with u-substitution, you have everything you need to start here.
The integral of 1/u with respect to u is ln|u| + C. Most of the work in this section is about recognising when a given integral is secretly in that form, using log properties to simplify, and knowing the standard trig integrals that produce logarithms.
Natural logarithm (ln x)
The logarithm with base e (Euler's number, approximately 2.718). It is the inverse of the exponential function e^x. In simple terms, ln x answers the question: "e raised to what power gives x?"
Logarithmic properties
The algebraic rules that let you break apart or combine logarithmic expressions: product rule, quotient rule, and power rule. Think of these as the tools you use to simplify a messy log expression before differentiating or integrating.
u-substitution (in the context of ln integration)
The technique of letting u equal the denominator (or the "inside" function), computing du, and rewriting the integral as ∫ (du/u) = ln|u| + C. In simple terms, you're pattern-matching the integral into a form where the natural log pops out.
These are essential for simplifying expressions before differentiating or integrating:
ln(1) = 0
ln(ab) = ln a + ln b (product rule)
ln(a^n) = n · ln a (power rule)
ln(a/b) = ln a – ln b (quotient rule)
You will use these constantly when doing logarithmic differentiation (covered in Part 2).
The foundational integral:
∫ (1/x) dx = ln|x| + C
In u-substitution form:
∫ (du/u) = ln|u| + C
The absolute value bars matter. Since ln is only defined for positive inputs, the absolute value ensures the antiderivative works for negative x values as well.
Example 1: Simple constant multiple
∫ (2/x) dx = 2 ∫ (1/x) dx = 2 ln|x| + C
Pull the constant out, apply the rule directly.
Example 2: u-substitution with a linear denominator
∫ (1/(4x – 1)) dx
Let u = 4x – 1, so du = 4 dx, meaning dx = du/4
The integral becomes (1/4) ∫ (du/u)
Result: (1/4) ln|4x – 1| + C
The key move is multiplying and dividing by the derivative of the inner function.
Example 3: Long division before integrating
∫ (x² + x + 1)/(x² + 1) dx
The degree of the numerator equals the degree of the denominator, so you need polynomial long division first:
(x² + x + 1) ÷ (x² + 1) = 1 remainder x
Rewrite as: ∫ 1 dx + ∫ x/(x² + 1) dx
The second integral uses u = x² + 1, du = 2x dx
Result: x + (1/2) ln|x² + 1| + C
When the numerator's degree is greater than or equal to the denominator's, always do the division first.
These are standard results worth memorising:
∫ tan x dx = –ln|cos x| + C
∫ sec u du = ln|sec u + tan u| + C
∫ csc u du = –ln|csc u + cot u| + C
Each of these comes from a u-substitution where the result is a natural log. They appear frequently on exams and in later coursework.
Integral | Result |
|---|---|
∫ (1/x) dx | ln|x| + C |
∫ (du/u) | ln|u| + C |
∫ tan x dx | –ln|cos x| + C |
∫ sec x dx | ln|sec x + tan x| + C |
∫ csc x dx | –ln|csc x + cot x| + C |
The integral of 1/x appears constantly in physics and engineering. Radioactive decay, population models, and electrical circuits with capacitors all involve integrals that resolve to natural logs. Any time a rate of change is proportional to the reciprocal of a quantity, you end up here.
Forgetting the absolute value bars. Students write ln(x) instead of ln|x| in their antiderivatives. The absolute value is required because the integrand 1/x is defined for negative x, but ln(x) is not.
Trying to use the power rule on 1/x. You cannot write ∫ x^(–1) dx = x^0 / 0, because division by zero is undefined. This is the entire reason the ln rule exists as a separate case.
Skipping long division. When the numerator degree ≥ denominator degree, students try to force a u-substitution that does not work. Always divide first.
Dropping the constant of integration. On free-response exam questions, missing "+ C" on an indefinite integral costs marks every time.
⚠️ The integral ∫ (1/x) dx = ln|x| + C is tested in nearly every unit exam and on the AP exam. Know it cold.
⚠️ Recognising when to use long division before integrating is a common free-response setup.
⚠️ The three trig integrals (tan, sec, csc) are frequently tested in multiple-choice sections. Memorise them.
⚠️ Always include absolute value bars around the argument of ln in an antiderivative.
True or false: ∫ (1/x) dx = ln(x) + C.
A: False. It should be ln|x| + C (absolute value required).
Fill in the blank: ∫ (du/u) = ______
A: ln|u| + C
True or false: To integrate (x² + 1)/(x + 1), you should try u-substitution first.
A: False. The numerator's degree is higher, so polynomial long division comes first.
Q: Evaluate ∫ (5/x) dx.
A: 5 ln|x| + C. Pull the constant 5 out front and apply the standard rule.
Q: Evaluate ∫ (1/(3x + 7)) dx.
A: (1/3) ln|3x + 7| + C. Let u = 3x + 7, du = 3 dx, so the integral is (1/3) ∫ du/u.
Q: Evaluate ∫ tan x dx.
A: –ln|cos x| + C. This is a standard result (equivalently, ln|sec x| + C).
Q: Why can't you use the power rule to integrate x^(–1)?
A: The power rule gives x^(n+1)/(n+1), and when n = –1 the denominator is zero. The natural log rule handles this special case instead.
Q: Evaluate ∫ (x² + 3)/(x² + 1) dx.
A: Perform long division first: (x² + 3)/(x² + 1) = 1 + 2/(x² + 1). The integral becomes x + 2 arctan(x) + C. (Note: this one produces arctan, not ln, showing why you must divide first to see the true form.)
This material connects directly to u-substitution (Unit 4), since nearly every ln integral requires spotting the substitution. It also lays the groundwork for logarithmic differentiation (Part 2 of these notes) and for integration techniques in Calc BC, where partial fractions produce ln terms. The trig integrals here reappear in trigonometric substitution problems.
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