Antiderivatives, logarithms and inverse circular forms: VCE Specialist Mathematics Units 3 & 4

1. Reverse a relationship, not just a symbol

An antiderivative of f is a function F for which F′(x) = f(x) throughout an interval. The indefinite integral ∫f(x) dx names the family F(x) + C. Differentiating loses a constant, so reversing differentiation cannot recover the starting height without extra information. The curves y = x², y = x² + 4 and y = x² − 3 all have gradient 2x at the same x-coordinate. They are vertical translations, not different solutions to the gradient relationship.

The word interval matters. If two differentiable functions have the same derivative on one connected interval, their difference is constant there. For f(x) = 1/x, the negative and positive half-lines are separate intervals. A condition at x = 2 determines the constant on the positive half-line but tells us nothing about an unrelated branch on the negative half-line. Never draw an antiderivative through an excluded point just because a calculator joins its plotted samples.

A derivative graph tells you how every member of the family behaves: f > 0 means F increases; f < 0 means F decreases. A sign change from negative to positive gives a local minimum of F. Adding C changes the height of that minimum, not its x-coordinate. This connection makes graphical reasoning a useful independent check on algebraic integration.

2. Logarithms belong to intervals

The power rule ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C requires n ≠ −1. At n = −1 the denominator would be zero, so a different derivative relationship is needed. For real x ≠ 0, the derivative of ln|x| is 1/x. When x is positive, this is the familiar derivative of ln x. When x is negative, ln|x| = ln(−x), and the chain rule gives (−1)/(−x) = 1/x.

The absolute-value bars make the logarithm argument positive; they do not turn the integrand into 1/|x|. These functions have different signs when x is negative. Likewise, ln|x² − 4| is meaningful on each interval separated by −2 and 2, but its derivative is undefined at those boundary points. Find the original exclusions before simplifying your answer.

For a linear denominator ax + b with a ≠ 0, the numerator needed for a direct logarithm is a. Therefore ∫1/(ax + b) dx = (1/a)ln|ax + b| + C. That factor comes from the chain rule. Dividing by a variable inside derivative is not a general shortcut: ∫1/(x² + 1) dx is not ln(x² + 1)/(2x).

3. Recognise inverse circular forms

Use a positive scale parameter a. The two central patterns are a difference of squares inside a square root and a sum of squares without a square root. Their antiderivatives have different inverse circular functions and different factors. The notation arcsin and arctan avoids confusing inverse functions with reciprocals such as cosec and cot.

Here arcsin has range [−π/2, π/2], while arctan has range (−π/2, π/2). These branch choices make each inverse a function. In the first formula the integrand is undefined at ±a even though arcsin itself has endpoint values. Our ordinary definite-integral calculations use bounds within the integrand’s continuous interval. Do not silently treat endpoint singularities as routine applications of the fundamental theorem.

To verify the first formula, differentiate arcsin(x/a): the chain rule gives (1/a)/√(1 − x²/a²). Because a > 0, √(a²) = a, so this simplifies to 1/√(a² − x²). In the second formula the outer factor 1/a cancels the extra scale arising from 1 + x²/a². Deriving these factors is safer than trusting visual similarity.

Complete the square or factor a positive constant when a denominator is disguised. For example, x² + 4x + 13 = (x + 2)² + 9 points towards arctan((x + 2)/3). In contrast, a numerator proportional to 2x + 4 points towards a logarithm. The denominator alone does not choose the method; compare the numerator with its derivative as well.

4. A definite integral measures net change

If f is continuous on [a, b] and F′ = f, then ∫_a^b f(x) dx = F(b) − F(a). Any constant cancels. The result is a number, whereas an indefinite integral describes functions. Brackets around the complete lower-bound expression prevent a subtraction error, especially when that expression contains several terms or a negative logarithm.

If f is a rate, the integral is accumulated change. A positive rate gives increasing quantity; a negative rate gives decreasing quantity. If both signs occur, signed contributions can cancel. Geometric area instead uses non-negative strip heights and may require splitting the interval. When time is in minutes and a rate is in litres per minute, multiplying rate by time gives litres, exactly the units expected for the integral.

5. Make checking a habit

Before integrating, name the variable and state any excluded values. Decide whether a power, logarithm, inverse sine or inverse tangent derivative genuinely matches. After integrating, differentiate the entire candidate, including inner functions and numerical factors. Only then use a condition or substitute bounds. Check the final units and whether a signed or non-negative result was requested.

When technology is available, compare the derivative of your candidate with the original expression using simplification on the stated domain. A numerical test at several safe points can expose a sign error but cannot prove an identity. A computer algebra system may return −arccos(x/a) rather than arcsin(x/a); on the common real interval these differ by a constant. Different appearance is not evidence that one is wrong.

A useful non-routine question is whether initial information really determines the requested function. A value at an excluded point cannot be imposed. Two inconsistent values may admit no solution. Values on disconnected intervals can determine separate constants. These questions use the meaning of an antiderivative rather than a new integration rule, and are especially valuable in linked modelling problems.

6. Separate scale, branch and starting information

A more general recognition pattern helps you organise a transformed inverse circular integral. With a > 0 and b ≠ 0, differentiating arctan((bx + c)/a) gives ab/[a² + (bx + c)²]. Therefore a numerator k requires the factor k/(ab). For the square-root pattern, differentiating arcsin((bx + c)/a) gives b/√[a² − (bx + c)²], so the required factor is k/b. These results follow from the chain rule, including when b is negative.

Do not memorise those factors independently of their domains. The arctangent expression is real for every real x, while the inverse-sine integrand needs |bx + c| < a. Solve that inequality carefully if b is negative. If b = 0, the proposed general formulas divide by zero; instead check whether the original integrand is a real constant and integrate it directly.

Initial values are also information to test, rather than instructions to substitute blindly. If a rate is positive throughout an interval, a compatible quantity must increase between any two ordered times in it. Two supplied values that decrease would be inconsistent with that rate. This graphical observation can expose a transcription or modelling error before you solve for C.

In a physical logarithmic model, ratios such as ln(t/t₀) keep the logarithm argument dimensionless, with t and t₀ measured in the same unit. Changing the reference value t₀ shifts the antiderivative by a constant; it does not change its derivative. This connects the apparently arbitrary constant with a concrete choice of reference level.