Integrating 1/x & Logarithmic Forms
Learn integrating \(\dfrac{1}{x}\) and logarithmic forms for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-05). The power rule fails for \(x^{-1}\), and the natural logarithm fills that gap: the primitive of \(\dfrac{1}{x}\) is \(\ln|x|\) plus a constant.
You will learn to integrate \(\dfrac{1}{x}\) and \(\dfrac{1}{ax + b}\), recognise fractions where the top is the derivative of the bottom, use absolute value signs correctly and evaluate definite logarithmic integrals — closing the one gap left by the power rule.
Theory
\(\int\dfrac1x dx=\ln|x|+C\) — the case the power rule cannot do. This Year 12 Mathematics Advanced topic (MAV-12-05) integrates \(\dfrac1x\) and \(\dfrac{f'}{f}\) forms.
\(\displaystyle\int\dfrac1x dx=\ln|x|+C\). More generally \(\displaystyle\int\dfrac{f'(x)}{f(x)}dx=\ln|f(x)|+C\) — when the top is the derivative of the bottom.
For a linear bottom, \(\displaystyle\int\dfrac{1}{ax+b}dx=\dfrac1a\ln|ax+b|+C\).
Method
- Recognise a fraction whose top is (a multiple of) the derivative of the bottom.
- Write \(\ln|\text{bottom}|\).
- Divide by \(a\) for a linear bottom.
| \(\ \) | \(=\) | \(\ln|x|+C\) |
| \(\ \) | \(=\) | \(\dfrac12\ln|2x+1|+C\) |
| \(\ \) | \(=\) | \(\ln(x^{2}+1)+C\) |
| \(\ \) | \(=\) | \(\big[\ln x\big]_1^{e}=1\) |
Common pitfalls
Frequently asked questions
What is the integral of 1 over x?
It is ln of the absolute value of x, plus C.
Why is there an absolute value in ln mod x?
Because the logarithm is only defined for positive numbers, the absolute value lets the rule work for negative x as well.
How do you integrate 1 over (2x+1)?
Divide by the coefficient 2: the integral is one half ln of the absolute value of 2x+1, plus C.
When does an integral give a logarithm?
When the integrand is a fraction whose numerator is the derivative of its denominator, the integral is ln of the absolute value of the denominator.