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)}$1/f(x), the derivative of ln at f(x), times f'(x), the derivative of the inside. So the derivative of ln f(x) is $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$f'(x)/f(x): a fraction whose numerator is the derivative of its denominator.

For example, $\ln \left(x^2+1\right)$ln(x² + 1) differentiates to $\frac{2x}{x^2+1}$2x/(x² + 1), and ln(sin x) to $\frac{\cos x}{\sin x}$(cos x)/(sin x).

Read backwards

Reading that the other way gives an integral: $\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\ln |f\left(x\right)|+C$∫ f'(x)/f(x) dx = ln|f(x)| + 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 $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$f'(x)/f(x).

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 $\int \frac{2x}{x^2+1}dx$∫ 2x/(x² + 1) dx. The denominator $x^2+1$x² + 1 differentiates to 2x, which is the numerator, so the integral is $\ln \left(x^2+1\right)+C$ln(x² + 1) + C. No absolute value is needed, since $x^2+1$x² + 1 is always positive.

Definite integrals follow. From 1 to 3 the value is $\ln 10-\ln 2=\ln 5\approx 1.609$ln 10 − ln 2 = ln 5 ≈ 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 $y=\frac{2x}{x^2+1}$y = 2x/(x² + 1), shaded from x = 0 to x = 2. The plain curve is $y=\ln \left(x^2+1\right)$y = ln(x² + 1), which starts at 0. Its height at x = 2 is $\ln 5\approx 1.609$ln 5 ≈ 1.609, the shaded area, because the area from 0 up to any x is $\ln \left(x^2+1\right)-\ln 1$ln(x² + 1) − 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}$3x²/(x³ + 5) the denominator differentiates to $3x^2$3x², the numerator exactly, so the integral is $\ln |x^3+5|+C$ln|x³ + 5| + C.

A numerator that is a constant multiple of the derivative works too, with the constant fixed outside. In $\frac{x}{x^2+4}$x/(x² + 4), the denominator differentiates to 2x, and x is ½ of 2x. So $\int \frac{x}{x^2+4}dx=\frac{1}{2}\int \frac{2x}{x^2+4}dx=\frac{1}{2}\ln \left(x^2+4\right)+C$∫ x/(x² + 4) dx = ½ ∫ 2x/(x² + 4) dx = ½ ln(x² + 4) + C. The integral from 0 to 2 is $\frac{1}{2}\left(\ln 8-\ln 4\right)=\frac{1}{2}\ln 2\approx 0.347$½(ln 8 − ln 4) = ½ ln 2 ≈ 0.347.

In the same way, $x^3+1$x³ + 1 differentiates to $3x^2$3x², and $x^2$x² is a third of that: $\int \frac{x^2}{x^3+1}dx=\frac{1}{3}\ln |x^3+1|+C$∫ x²/(x³ + 1) dx = ⅓ ln|x³ + 1| + C. From 0 to 1 that is $\frac{1}{3}\ln 2\approx 0.231$⅓ ln 2 ≈ 0.231.

The pattern in other families

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

tan x is $\frac{\sin x}{\cos x}$(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: $\int \tan xdx=-\ln |\cos x|+C$∫ tan x dx = −ln|cos x| + C.

When it does not fit

Only a constant factor can be adjusted. In $\frac{x}{x^3+1}$x/(x³ + 1), the denominator differentiates to $3x^2$3x², and x is not a constant times $x^2$x², 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}$1/(x² + 1), the denominator differentiates to 2x, and the numerator 1 has no x at all. That integral is ${\tan }^{-1}x+C$tan⁻¹x + C, not a logarithm. A denominator that factors, such as $x^2-1$x² − 1 with a numerator of 1, is split into partial fractions instead.

The usual mistakes

Answering $\frac{1}{f}$1/f. $\frac{1}{x^2+1}$1/(x² + 1) differentiates to $-\frac{2x}{{\left(x^2+1\right)}^2}$−2x/(x² + 1)², which has a squared denominator; the integral of $\frac{2x}{x^2+1}$2x/(x² + 1) is $\ln \left(x^2+1\right)+C$ln(x² + 1) + C.

Taking the log of the numerator. ln|cos x| differentiates to $-\frac{\sin x}{\cos x}$−(sin x)/(cos x), not $\frac{\cos x}{\sin x}$(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: $\frac{1}{2}\ln \left(x^2+4\right)+C$½ ln(x² + 4) + C.

Forcing the pattern where a variable is missing. Any multiple $k\ln |x^3+1|$k ln|x³ + 1| differentiates to $\frac{3k{x}^2}{x^3+1}$3kx²/(x³ + 1), with $x^2$x² on top, never x, so no multiple of $\ln |x^3+1|$ln|x³ + 1| is the integral of $\frac{x}{x^3+1}$x/(x³ + 1).