Anti-differentiation by recognition (reverse chain rule): VCE Math Methods Unit 4

What this dot point is asking

The VCE Mathematics Study Design (from 2023) lists, for Mathematical Methods Units 3 and 4, "anti-differentiation by recognition that $F'(x) = f(x)$ implies $\int f(x) \, dx = F(x) + c$". In plain terms: if you can see (or are told) that some function $F$ differentiates to the integrand, then $F$ is an anti-derivative. The other calculus key knowledge you use here is the list of standard anti-derivatives, including functions of the form $f(ax + b)$ (a linear inside function).

Formal integration by substitution, where you write $u = g(x)$ and change the variable, is not Methods content. The same study design lists "use of the substitution $u = g(x)$ to anti-differentiate expressions" under Specialist Mathematics Units 3 and 4. So in Methods, an integral with a non-linear inside function, such as $\int 2x(x^2 + 1)^3 \, dx$, is almost always set up so you can recognise the answer, typically after a "hence" part that asks you to differentiate something first.

Linear inside functions: the $f(ax + b)$ rule

When the inside function is linear, there is a direct rule. If $F$ is an anti-derivative of $f$, then

The $\frac{1}{a}$ undoes the factor of $a$ that the chain rule would produce on differentiating.

• $\int (3x + 1)^4 \, dx = \frac{1}{3} \cdot \frac{(3x + 1)^5}{5} + c = \frac{(3x + 1)^5}{15} + c$.
• $\int \cos(3x) \, dx = \frac{1}{3}\sin(3x) + c$.
• $\int \sqrt{2x + 1} \, dx = \frac{1}{2} \cdot \frac{(2x + 1)^{3/2}}{3/2} + c = \frac{1}{3}(2x + 1)^{3/2} + c$.
• $\int \frac{1}{(2x + 1)^2} \, dx = \frac{1}{2} \cdot \frac{(2x + 1)^{-1}}{-1} + c = -\frac{1}{2(2x + 1)} + c$.

Non-linear inside functions: recognise the reverse chain rule

The chain rule says $\frac{d}{dx} F(g(x)) = F'(g(x)) \, g'(x)$. Read backwards: whenever the integrand looks like (a function of $g(x)$) multiplied by $g'(x)$, the anti-derivative is the matching function of $g(x)$. Methods questions usually make this explicit with a two-part structure.

The "hence" procedure

1. Differentiate the function the question gives you (or the natural candidate, such as the bracket raised to one higher power).
2. Compare the derivative with the integrand. They usually differ by a constant factor, or by extra terms.
3. Rearrange so the integrand is expressed in terms of the derivative.
4. Anti-differentiate using the fact that an anti-derivative of $F'(x)$ is $F(x)$.
5. Check by differentiating your answer. It takes seconds and catches a wrong constant.

Example 1. Power of a bracket

Find $\int 6x(x^2 + 1)^4 \, dx$.

Try $F(x) = (x^2 + 1)^5$. Then $F'(x) = 5(x^2 + 1)^4 \cdot 2x = 10x(x^2 + 1)^4$.

The integrand is $\frac{6}{10} F'(x) = \frac{3}{5}F'(x)$, so $\int 6x(x^2 + 1)^4 \, dx = \frac{3}{5}(x^2 + 1)^5 + c$.

Example 2. Exponential of a function

Find $\int x e^{x^2} \, dx$.

$\frac{d}{dx} e^{x^2} = 2x e^{x^2}$, so the integrand is half of this derivative: $\int x e^{x^2} \, dx = \frac{1}{2} e^{x^2} + c$.

Example 3. Logarithm of a function

Find an anti-derivative of $\frac{2x + 3}{x^2 + 3x + 5}$.

$\frac{d}{dx}\log_e(x^2 + 3x + 5) = \frac{2x + 3}{x^2 + 3x + 5}$. The quadratic has discriminant $9 - 20 < 0$ and a positive leading coefficient, so it is positive for all $x$ and the logarithm is defined everywhere. An anti-derivative is $\log_e(x^2 + 3x + 5)$.

Example 4. Trigonometric

Find $\int \sin^3(x) \cos(x) \, dx$.

$\frac{d}{dx}\sin^4(x) = 4\sin^3(x)\cos(x)$, so $\int \sin^3(x) \cos(x) \, dx = \frac{1}{4}\sin^4(x) + c$.

Example 5. A product-rule "hence"

Let $f(x) = x\log_e(x)$ for $x > 0$. Find $f'(x)$ and hence find $\int \log_e(x) \, dx$.

$f'(x) = \log_e(x) + 1$, so $\log_e(x) = f'(x) - 1$. Anti-differentiating, $\int \log_e(x) \, dx = x\log_e(x) - x + c$.

This is the pattern Methods uses to reach anti-derivatives that are not on the standard list: the question gives you a function whose derivative contains the integrand, and you isolate it.

Definite integrals by recognition

Once you have an anti-derivative, apply $\int_a^b f(x) \, dx = F(b) - F(a)$ as usual. There is no change of limits in Methods, because there is no change of variable.

$\int_{0}^{1} 2x (x^2 + 1)^3 \, dx = \left[ \frac{(x^2 + 1)^4}{4} \right]_{0}^{1} = \frac{2^4}{4} - \frac{1^4}{4} = \frac{15}{4}$.

Where $u$-substitution fits

If you also study Specialist Mathematics, you will meet the same integrals written with $u = g(x)$, $du = g'(x)\,dx$, and a change of limits. The answers agree, because substitution is just a formal way of organising the reverse chain rule. For Methods, the recognition method above is what the study design lists, and Methods questions with a non-linear inside function normally supply the derivative you need, so you should never be stuck without a substitution.

Examples in context

Example 1. Total accumulation. A pollutant enters a lake at rate $r(t) = 3t^2 e^{t^3}$ grams per hour. Since $\frac{d}{dt} e^{t^3} = 3t^2 e^{t^3}$, the total entering over the first hour is $\int_0^1 3t^2 e^{t^3}\,dt = \left[e^{t^3}\right]_0^1 = e - 1 \approx 1.72$ grams.

Example 2. A logarithmic rate. A rate is $\frac{2x}{x^2 + 1}$ units per unit of $x$. Since $\frac{d}{dx}\log_e(x^2 + 1) = \frac{2x}{x^2 + 1}$, the accumulated change from $x = 0$ to $x = 1$ is $\log_e(2) - \log_e(1) = \log_e(2)$.

Try this

Q1. Find $\frac{d}{dx}(x^2 + 3)^5$ and hence find $\displaystyle\int 2x(x^2 + 3)^4\,dx$. [3 marks]

• Cue. $\frac{d}{dx}(x^2 + 3)^5 = 10x(x^2 + 3)^4$, so $\int 2x(x^2 + 3)^4\,dx = \frac{(x^2 + 3)^5}{5} + c$.

Q2. Evaluate $\displaystyle\int_0^1 x e^{x^2}\,dx$. [2 marks]

• Cue. $\frac{d}{dx}e^{x^2} = 2xe^{x^2}$, so the value is $\frac{1}{2}\left[e^{x^2}\right]_0^1 = \frac{1}{2}(e - 1)$.

Q3. Differentiate $\sin^3(x)$ and hence find $\displaystyle\int \cos(x) \sin^2(x)\,dx$. [3 marks]

• Cue. $\frac{d}{dx}\sin^3(x) = 3\sin^2(x)\cos(x)$, so the integral is $\frac{1}{3}\sin^3(x) + c$.

Q4. Find $\displaystyle\int (4x - 1)^{-3}\,dx$. [2 marks]

• Cue. Linear inside: $\frac{1}{4} \cdot \frac{(4x - 1)^{-2}}{-2} + c = -\frac{1}{8(4x - 1)^2} + c$.