Before the method: the idea behind it

Filling the hole in the power rule

There has been a gap in your integration since the very first lesson on anti-differentiation, hidden behind a small condition written in brackets. This lesson fills it, and the function that fills it is the natural logarithm. That is not a coincidence: it turns out the logarithm can be defined as an area, which is where its calculus comes from.

Part one

The one case the rule could not handle

The reverse power rule says to raise the power by one and divide by the new power. It works for every power except one.

\[ \int x^n\,dx=\frac{x^{n+1}}{n+1}+c \qquad (n\ne-1) \]

That condition in brackets has been there all along. Look at what happens when n = −1: raising the power by one gives x⁰, and dividing by the new power means dividing by zero. The rule collapses.

the integral	raise the power	divide by	result
∫x³ dx	x⁴	4	x⁴/4
∫x dx	x²	2	x²/2
∫x⁰ dx	x¹	1	x
∫x⁻¹ dx	x⁰	0	undefined
∫x⁻² dx	x⁻¹	−1	−1/x

Every row works except one. So what is the anti-derivative of 1/x?

The answer cannot be any power of x, because the power rule would have found it. It has to be a completely different kind of function, and by the fundamental theorem it must be the area function under y = 1/x.

Part two

The area under 1/x is a logarithm

Sweep out the area under y = 1/x, starting from x = 1, and watch what the accumulated area does.

Upper limit: x = 1.00

Area so far: 0.000

ln x: 0.000

Watch the two readouts at the bottom. They are always identical.

The area and ln x match at every point, and that is not an accident of this particular curve. The area starts at zero when x = 1, exactly as ln 1 = 0. It grows without bound but ever more slowly, exactly as ln x does. It reaches 1 when x = e, which is precisely the property that defines e.

Historically this is how the natural logarithm was first understood: not as an exponent, but as an area. The two descriptions turn out to be the same function, and it is that fact which supplies the missing anti-derivative.

\[ \dfrac{d}{dx}\left(\ln x\right)=\dfrac{1}{x}, \qquad \int\dfrac{1}{x}\,dx=\ln x+c \quad (x \gt 0) \]

The restriction x > 0 is there because ln is only defined for positive numbers. Its consequence in practice: after computing a definite integral of 1/x, always check that the interval does not cross zero, because the function is not even defined there.

Part three

When the inside is a function

Combine the new derivative with the chain rule and a very useful pattern appears.

Differentiating ln(f(x)) gives 1 over f(x), multiplied by the derivative of the inside. So the result is always the inside’s derivative divided by the inside itself.

\[ \dfrac{d}{dx}\Big(\ln f(x)\Big)=\dfrac{f'(x)}{f(x)}, \qquad\text{and in reverse}\qquad \int\dfrac{f'(x)}{f(x)}\,dx=\ln\big|f(x)\big|+c \]

The reverse direction is the one worth training your eye for. Whenever you meet a fraction whose numerator is the derivative of its denominator, the integral is simply the natural logarithm of that denominator. Spotting this pattern turns integrals that look impossible into one-line answers.

For example, in \(\displaystyle\int\frac{2x}{x^2+1}\,dx\), the numerator 2x is exactly the derivative of the denominator x² + 1, so the answer is \(\ln(x^2+1)+c\) with no working required beyond noticing.

If the numerator is a constant multiple off, adjust it. For \(\displaystyle\int\frac{x}{x^2+4}\,dx\) the derivative of the bottom is 2x, and the top is half of that, so the answer is \(\frac{1}{2}\ln(x^2+4)+c\).

Where this shows up in real life

Anything with a 1/x rate

When a rate of change is inversely proportional to the quantity already present, integrating gives a logarithm. This describes how quickly a diluting solution loses concentration, and how the marginal value of extra resources declines as you accumulate more of them.

Information and entropy

The amount of information in a message is measured with logarithms, which is why file sizes are quoted in bits. The mathematics of data compression rests directly on the calculus of logarithmic functions.

Before you start

What to reach for

The question involves…	Use
differentiating ln of something	f′(x) over f(x)
ln multiplied by or divided by another function	Product or quotient rule, with 1/x as the log’s derivative
integrating 1/x or 1/(ax + b)	ln of the denominator, divided by a
a fraction whose top is the derivative of its bottom	ln of the bottom, adjusting any constant factor
ln of a power or product	Simplify with the log laws first, then differentiate

Next: the working method, twenty practice questions, ten harder ones, and the ClassPad guide.

Start with the method →