To integrate a reciprocal of a linear expression:

Why? This follows from substitution: let \(u = ax+b\).

Let \(u = ax + b\), so \(du = a\,dx\) → \(dx = \frac{du}{a}\).

\[ \int \frac{1}{ax+b}\,dx = \int \frac{1}{u}\cdot \frac{du}{a} = \frac{1}{a}\int \frac{1}{u}du = \frac{1}{a}\ln|u| + C = \frac{1}{a}\ln|ax+b| + C \]

Example 1: \(\displaystyle \int \frac{1}{3x+7}\,dx\)

\[ \int \frac{1}{3x+7}\,dx = \frac{1}{3}\ln|3x+7| + C \]

Example 2: \(\displaystyle \int \frac{5}{2x-1}\,dx\)

Factor out the constant: \[ 5\int\frac{1}{2x-1}\,dx = 5\cdot \frac{1}{2}\ln|2x-1| + C = \frac{5}{2}\ln|2x-1| + C \]

Example 3: \(\displaystyle \int \frac{3x+4}{x+2}\,dx\)

Split into parts: \[ \frac{3x+4}{x+2} = 3 + \frac{-2}{x+2} \] So: \[ \int 3\,dx - 2\int\frac{1}{x+2}dx = 3x - 2\ln|x+2| + C \]

Example: \(\displaystyle \int_{0}^{2} \frac{1}{4x+1}\,dx\)

\[ \int \frac{1}{4x+1}\,dx = \frac{1}{4}\ln|4x+1| \] Apply limits: \[ \left.\frac{1}{4}\ln|4x+1|\right|_{0}^{2} = \frac{1}{4}\left(\ln|9| - \ln|1|\right) = \frac{1}{4}\ln 9 \]

• Always check if you can factor out a constant first.
• Use partial fractions if the expression is more complicated than a single linear term.
• Always include absolute values inside the logarithm.
• In definite integrals, do not substitute back if limits have been changed to \(u\).
• Cambridge often combines this form with partial fractions.