The use of substitution to evaluate integrals of the form $\int f(g(x)) g'(x) \, dx$, recognising the reverse of the chain rule

What this dot point is asking

VCAA wants you to recognise integrals of the form $\int f(g(x)) g'(x) \, dx$ (composed function multiplied by the derivative of the inside) and evaluate them using the substitution method. The dot point is the integration analogue of the chain rule. Paper 1 typically includes one substitution question every year.

The reverse chain rule

Recall the chain rule for differentiation. If $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

Reversing this for integration: if the integrand has the structure $f'(g(x)) \cdot g'(x)$, the antiderivative is $f(g(x))$.

Substitution is the systematic procedure for spotting and unwinding this structure.

The substitution procedure

To evaluate $\int f(g(x)) g'(x) \, dx$ (or any integral where one part is the derivative of another part):

1. Let $u = g(x)$. Choose $u$ to be the inside function.
2. Differentiate to get $\frac{du}{dx} = g'(x)$, then write $du = g'(x) \, dx$.
3. Substitute $u$ and $du$ into the integral. The $x$-variables should all disappear; the integral becomes an integral in $u$.
4. Evaluate the simpler integral.
5. For indefinite integrals, replace $u$ with $g(x)$ at the end. For definite integrals, change the limits to $u$ values (or substitute back to $x$ before evaluating).

The choice of $u$ is the key step. Look for the inside function whose derivative appears (possibly with a constant factor) outside as part of the integrand.

Indefinite integrals: examples

Example 1. Power inside a polynomial

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

Choose $u = x^2 + 1$. Then $du = 2x \, dx$, so $6 x \, dx = 3 \, du$.

Rewrite. $\int 3 u^4 \, du = 3 \cdot \frac{u^5}{5} + c = \frac{3 u^5}{5} + c$.

Substitute back. $\frac{3 (x^2 + 1)^5}{5} + c$.

Example 2. Exponential of a function

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

Choose $u = x^2$. Then $du = 2x \, dx$, so $x \, dx = \frac{1}{2} du$.

Rewrite. $\int \frac{1}{2} e^{u} \, du = \frac{1}{2} e^{u} + c$.

Substitute back. $\frac{1}{2} e^{x^2} + c$.

Example 3. $\frac{1}{x}$-style logarithmic

$\int \frac{2x + 3}{x^2 + 3x + 5} \, dx$.

Choose $u = x^2 + 3x + 5$. Then $du = (2x + 3) \, dx$.

Rewrite. $\int \frac{1}{u} \, du = \ln\lvert u \rvert + c$.

Substitute back. $\ln\lvert x^2 + 3x + 5 \rvert + c$.

Example 4. Trigonometric

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

Choose $u = \sin(x)$. Then $du = \cos(x) \, dx$.

Rewrite. $\int u^3 \, du = \frac{u^4}{4} + c$.

Substitute back. $\frac{\sin^4(x)}{4} + c$.

Definite integrals: limits in IMATH_42

For definite integrals, you have two options.

Option A. Change the limits. When $u = g(x)$, the lower limit $x = a$ becomes $u = g(a)$ and the upper limit $x = b$ becomes $u = g(b)$. Evaluate the integral in $u$ directly between the new limits. No need to substitute back.

Option B. Substitute back. Compute the antiderivative in $u$, replace $u$ with $g(x)$, then evaluate at the original $x$-limits.

Both produce the same answer. Option A is usually faster.

Example. Option A

$\int_{0}^{1} 2x (x^2 + 1)^3 \, dx$.

Choose $u = x^2 + 1$, $du = 2x \, dx$.

Change limits. When $x = 0$, $u = 1$. When $x = 1$, $u = 2$.

Rewrite. $\int_{1}^{2} u^3 \, du = \left[ \frac{u^4}{4} \right]_{1}^{2} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}$.

Example. Option B (same integral)

Compute antiderivative in $u$. $\frac{u^4}{4} = \frac{(x^2 + 1)^4}{4}$.

Evaluate at original limits. $\left[ \frac{(x^2 + 1)^4}{4} \right]_{0}^{1} = \frac{2^4}{4} - \frac{1^4}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}$.

Same answer.

Recognising when substitution is needed

Three patterns to recognise:

1. Outside function is the derivative of the inside. $\int 2x (x^2 + 1)^3 \, dx$: the $2x$ outside is the derivative of $x^2 + 1$ inside.

2. Numerator is the derivative of the denominator. $\int \frac{f'(x)}{f(x)} \, dx = \ln\lvert f(x) \rvert + c$. The substitution $u = f(x)$ converts this to $\int \frac{1}{u} \, du$.

3. Constant-multiple version. Sometimes the derivative of the inside appears with the wrong coefficient. For example, $\int x (x^2 + 1)^3 \, dx$: the derivative of $x^2 + 1$ is $2x$, but only $x$ appears. Adjust with a constant: $x \, dx = \frac{1}{2} du$.

When substitution does not work

Substitution is the reverse chain rule. It does not apply when the integrand has no obvious inner / outer structure. Some integrals require:

• Linear reverse chain only (without substitution): $\int f(ax + b) \, dx = \frac{1}{a} F(ax + b) + c$. Faster than full substitution.
• Integration by parts (not in the VCE Methods Unit 4 syllabus).
• Partial fractions (also outside Methods).

If you cannot identify a clean $u$ whose $du$ matches part of the integrand, substitution may not be the right tool.

Common errors

Choosing $u$ as the wrong piece. The standard heuristic: $u$ is the inside function whose derivative appears in the integrand. Choosing $u = x$ or $u =$ the outside function rarely simplifies.

Forgetting to substitute $du$ for $dx$. $\int f(u) \cdot dx$ is mixed notation; the $dx$ must be converted to $du$ before integrating.

Forgetting to change the limits in definite integrals. If you choose Option A but use the original $x$-limits with the $u$-integrand, the arithmetic is wrong.

Not substituting back in indefinite integrals. $\frac{u^5}{5} + c$ is not a complete answer for an integral originally in $x$; replace $u$ with $g(x)$ before submitting.

Constants getting lost. If the derivative of the inside is $2x$ but only $x$ is in the integrand, you must compensate with a factor of $\frac{1}{2}$. Forgetting halves the answer.

Using substitution when the linear reverse chain rule would do. For $\int (3x + 1)^4 \, dx$, the reverse chain gives $\frac{(3x + 1)^5}{15} + c$ directly; full substitution works but is slower.

In one sentence

Integration by substitution reverses the chain rule by letting $u = g(x)$ for the inside function, computing $du = g'(x) \, dx$, rewriting the integral entirely in $u$, evaluating the simpler integral, and (for indefinite cases) substituting back to $x$; the central skill is choosing $u$ so that its derivative matches a factor already present in the integrand.