Integrals of f′ over f

The derivative of ln f

By the chain rule, ln f(x) differentiates to $\frac{1}{f\left(x\right)}$, the derivative of ln at f(x), times f'(x), the derivative of the inside. So the derivative of ln f(x) is $\frac{f'\left(x\right)}{f\left(x\right)}$: a fraction whose numerator is the derivative of its denominator.

For example, $\ln\left(x^{2} + 1\right)$ differentiates to $\frac{2x}{{x^{2} + 1}}$, and ln(sin x) to $\frac{{\cos x}}{{\sin x}}$.

Read backwards

Reading that the other way gives an integral: $\int \frac{f'\left(x\right)}{f\left(x\right)} \,dx = \ln|f\left(x\right)| + C$. The absolute value is there for the same reason as in ln|x|: f(x) may be negative, and ln|f(x)| still differentiates to $\frac{f'\left(x\right)}{f\left(x\right)}$.

The numerator has to be exactly the derivative of the denominator, or a constant multiple of it, as below. Anything else and the pattern does not apply.

Checking the numerator

Take $\int \frac{2x}{{x^{2} + 1}} \,dx$. The denominator $x^{2} + 1$ differentiates to 2x, which is the numerator, so the integral is $\ln\left(x^{2} + 1\right) + C$. No absolute value is needed, since $x^{2} + 1$ is always positive.

Definite integrals follow. From 1 to 3 the value is $\ln 10 - \ln 2 = \ln 5 \approx 1.609$. From 0 to 2 it is ln 5 − ln 1 = ln 5 as well, so the two regions have the same area.

The gold curve is $y = \frac{2x}{{x^{2} + 1}}$, shaded from x = 0 to x = 2. The plain curve is $y = \ln\left(x^{2} + 1\right)$, which starts at 0. Its height at x = 2 is $\ln 5 \approx 1.609$, the shaded area, because the area from 0 up to any x is $\ln\left(x^{2} + 1\right) - \ln 1$.

Adjusting a constant

The skill is the check: differentiate the denominator first, then compare it with the numerator. In $\frac{3x^{2}}{{x^{3} + 5}}$ the denominator differentiates to $3x^{2}$, the numerator exactly, so the integral is $\ln|x^{3} + 5| + C$.

A numerator that is a constant multiple of the derivative works too, with the constant fixed outside. In $\frac{x}{{x^{2} + 4}}$, the denominator differentiates to 2x, and x is ½ of 2x. So $\int \frac{x}{{x^{2} + 4}} \,dx = \frac{1}{2} \int \frac{2x}{{x^{2} + 4}} \,dx = \frac{1}{2} \ln\left(x^{2} + 4\right) + C$. The integral from 0 to 2 is $\frac{1}{2}\left(\ln 8 - \ln 4\right) = \frac{1}{2} \ln 2 \approx 0.347$.

In the same way, $x^{3} + 1$ differentiates to $3x^{2}$, and $x^{2}$ is a third of that: $\int \frac{x^{2}}{{x^{3} + 1}} \,dx = \frac{1}{3} \ln|x^{3} + 1| + C$. From 0 to 1 that is $\frac{1}{3} \ln 2 \approx 0.231$.

The pattern in other families

The denominator need not be a polynomial. sin x differentiates to cos x, so $\int \frac{{\cos x}}{{\sin x}} \,dx = \ln|\sin x| + C$. $e^{x} + 1$ differentiates to $e^{x}$, so $\int \frac{e^{x}}{{e^{x} + 1}} \,dx = \ln\left(e^{x} + 1\right) + C$.

tan x is $\frac{{\sin x}}{{\cos x}}$, and cos x differentiates to −sin x, so the numerator is minus the derivative of the denominator. The minus comes outside: $\int \tan x \,dx = -\ln|\cos x| + C$.

When it does not fit

Only a constant factor can be adjusted. In $\frac{x}{{x^{3} + 1}}$, the denominator differentiates to $3x^{2}$, and x is not a constant times $x^{2}$, so the pattern fails. The difference is a factor of x, and moving a variable outside the integral is not allowed.

In $\frac{1}{{x^{2} + 1}}$, the denominator differentiates to 2x, and the numerator 1 has no x at all. That integral is $\tan^{-1}x + C$, not a logarithm. A denominator that factors, such as $x^{2} - 1$ with a numerator of 1, is split into partial fractions instead.

The usual mistakes

Answering $\frac{1}{f}$. $\frac{1}{{x^{2} + 1}}$ differentiates to $- \frac{2x}{\left(x^{2} + 1\right)^{2}}$, which has a squared denominator; the integral of $\frac{2x}{{x^{2} + 1}}$ is $\ln\left(x^{2} + 1\right) + C$.

Taking the log of the numerator. ln|cos x| differentiates to $- \frac{{\sin x}}{{\cos x}}$, not $\frac{{\cos x}}{{\sin x}}$. The log is always of the denominator.

Fixing the constant the wrong way. The numerator x is ½ of 2x, so the answer carries ½, not 2: $\frac{1}{2} \ln\left(x^{2} + 4\right) + C$.

Forcing the pattern where a variable is missing. Any multiple $k \ln|x^{3} + 1|$ differentiates to $\frac{3kx^{2}}{{x^{3} + 1}}$, with $x^{2}$ on top, never x, so no multiple of $\ln|x^{3} + 1|$ is the integral of $\frac{x}{{x^{3} + 1}}$.