The reverse power rule fails for exactly one power: $x^{-1}$ would need division by 0. The derivative $\frac{d}{dx}\ln x = \frac{1}{x}$ from Derivatives of Exponential and Log Functions fills the gap, and Integration by Substitution extends it to a whole family of integrals.
Key idea
$$\int\frac{1}{x}\,dx = \ln\lvert x \rvert + C, \qquad \int e^x\,dx = e^x + C, \qquad \int a^x\,dx = \frac{a^x}{\ln a} + C,$$ and by substitution $\displaystyle\int\frac{g'(x)}{g(x)}\,dx = \ln\lvert g(x) \rvert + C$ (top = derivative of the bottom).
Why it works
For $x > 0$, $\frac{d}{dx}\ln x = \frac{1}{x}$. For $x < 0$, $\frac{d}{dx}\ln(-x) = \frac{-1}{-x} = \frac{1}{x}$ as well. The absolute value combines both cases, so $\ln\lvert x \rvert$ is an antiderivative of $\frac{1}{x}$ on each side of 0. Without the absolute value, the formula would be undefined for every negative $x$, where $\frac{1}{x}$ is perfectly happy.
Worked example: A linear bottom
Find $\displaystyle\int\frac{3}{2x + 1}\,dx$.
- $u = 2x + 1$, $du = 2\,dx$, so $dx = \frac{1}{2}du$.
- $\frac{3}{2}\int\frac{1}{u}\,du = \frac{3}{2}\ln\lvert u \rvert + C$.
Answer: $\frac{3}{2}\ln\lvert 2x + 1 \rvert + C$
Worked example: Top is the derivative of the bottom
Find $\displaystyle\int\frac{x}{x^2 + 4}\,dx$ and $\displaystyle\int\tan x\,dx$.
- $u = x^2 + 4$, $x\,dx = \frac{1}{2}du$: $\frac{1}{2}\ln(x^2 + 4) + C$ (no absolute value needed, since $x^2 + 4 > 0$).
- $\tan x = \frac{\sin x}{\cos x}$; with $u = \cos x$, $du = -\sin x\,dx$: $-\int\frac{du}{u} = -\ln\lvert\cos x\rvert + C$.
Answer: $\frac{1}{2}\ln(x^2 + 4) + C$ and $-\ln\lvert\cos x\rvert + C$
Worked example: Definite integrals
Evaluate $\displaystyle\int_1^{e}\frac{1}{x}\,dx$ and $\displaystyle\int_0^{\ln 3}e^x\,dx$.
- $\big[\ln\lvert x \rvert\big]_1^e = 1 - 0 = 1$.
- $\big[e^x\big]_0^{\ln 3} = 3 - 1 = 2$.
Answer: $1$ and $2$
| Integrand | Antiderivative |
|---|---|
| $e^{kx}$ | $\frac{1}{k}e^{kx} + C$ |
| $\frac{1}{ax + b}$ | $\frac{1}{a}\ln\lvert ax + b \rvert + C$ |
| $a^x$ | $\frac{a^x}{\ln a} + C$ |
| $\frac{g'(x)}{g(x)}$ | $\ln\lvert g(x) \rvert + C$ |
| $\tan x$ | $-\ln\lvert\cos x\rvert + C$ |
Common mistake
Two slips: dropping the absolute value in $\ln\lvert x \rvert$, and using the log rule where it does not belong. $\int\frac{1}{x^2}\,dx$ is not $\ln(x^2)$; it is $-\frac{1}{x} + C$ by the power rule. Check by differentiating: $\frac{d}{dx}\ln(x^2) = \frac{2}{x}$, which is not $\frac{1}{x^2}$.