1. Standard Integrals & Substitution

Video coming soon4 worked examples

Theory

Advanced Higher adds several standard integrals and the technique of substitution. Key results include:

\int \sec^2 x\,dx = \tan x + c, \quad \int \frac{1}{1+x^2}\,dx = \tan^{-1} x + c, \quad \int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c

Two further standard integrals come directly from reversing the inverse trigonometric derivatives. These appear on the formula sheet, but you must recognise when to reach for them:

\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \sin^{-1}\!\left(\frac{x}{a}\right) + c, \qquad \int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + c

The signal is the shape of the denominator: a square root of “constant minus $x^2$ ” points to $\sin^{-1}$ , while “constant plus $x^2$ ” with no root points to $\tan^{-1}$ . If the coefficient of $x^2$ is not $1$ , factor it out first to reach the standard form.

For integration by substitution, choose $u$ so that its derivative appears (up to a constant) in the integrand, and convert every part — including $dx$ — into $u$ .

The Golden Rule: spot the $\frac{f'(x)}{f(x)}$ and $g(f(x))f'(x)$ patterns for a quick substitution, and always replace $dx$ using $du = \frac{du}{dx}\,dx$ .

⚠️ Common Examiner Traps

Forgetting the $dx \to du$ step: the whole integrand, including $dx$ , must be in terms of $u$ .
Definite limits: either change the limits to $u$ -values or substitute back before applying them.
Missing the log pattern: $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c$ .

Worked examples

Example 1

Find $\displaystyle\int \frac{2x}{x^2 + 1}\,dx$ .

Step 1: The numerator is the derivative of the denominator, so this is the $\frac{f'}{f}$ pattern:

\frac{d}{dx}(x^2 + 1) = 2x

Step 2: Therefore:

\int \frac{2x}{x^2 + 1}\,dx = \ln|x^2 + 1| + c

Example 2

Find $\displaystyle\int 2x(x^2 + 1)^5\,dx$ using the substitution $u = x^2 + 1$ .

Step 1: With $u = x^2 + 1$ , we have $\frac{du}{dx} = 2x$ , so $du = 2x\,dx$ .

Step 2: Rewrite the integral in $u$ and integrate:

\int u^5\,du = \frac{u^6}{6} + c

Step 3: Substitute back:

= \frac{(x^2 + 1)^6}{6} + c

Example 3

Find $\displaystyle\int \frac{1}{9 + x^2}\,dx$ .

Step 1: This matches the standard form $\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\tan^{-1}\!\frac{x}{a} + c$ with $a = 3$ .

Step 2: Therefore:

\int \frac{1}{9 + x^2}\,dx = \frac{1}{3}\tan^{-1}\!\left(\frac{x}{3}\right) + c

Example 4

Find (a) $\displaystyle\int \frac{1}{\sqrt{25 - x^2}}\,dx$ and (b) $\displaystyle\int \frac{1}{9 + 4x^2}\,dx$ .

Step 1 (a): The denominator is a square root of “constant minus $x^2$ ”, so this is the $\sin^{-1}$ form with $a^2 = 25$ , giving $a = 5$ :

\int \frac{1}{\sqrt{25 - x^2}}\,dx = \sin^{-1}\!\left(\frac{x}{5}\right) + c

Step 2 (b): Here the coefficient of $x^2$ is not $1$ , so factor it out of the denominator first:

9 + 4x^2 = 4\left(\frac{9}{4} + x^2\right) \implies \int \frac{1}{9+4x^2}\,dx = \frac{1}{4}\int \frac{1}{\left(\frac{3}{2}\right)^2 + x^2}\,dx

Step 3: Now apply the standard form with $a = \dfrac{3}{2}$ :

= \frac{1}{4} \times \frac{1}{\frac{3}{2}}\tan^{-1}\!\left(\frac{x}{\frac{3}{2}}\right) + c = \frac{1}{4} \times \frac{2}{3}\tan^{-1}\!\left(\frac{2x}{3}\right) + c

Step 4: Simplify the constant:

\int \frac{1}{9+4x^2}\,dx = \frac{1}{6}\tan^{-1}\!\left(\frac{2x}{3}\right) + c