Integrating f'(x)/f(x) and Related Forms

When a fraction's numerator is the derivative of its denominator, or a constant multiple of it, the integral is a logarithm. This single pattern integrates $\dfrac{x}{x^2 + 1}$, $\dfrac{e^x}{e^x + 3}$, $\tan x$ and $\cot x$, and it supplies the logarithm terms in partial fraction integrals. The syllabus asks you to "recognise an integrand of the form $\dfrac{kf'(x)}{f(x)}$, and integrate such functions". This note also covers the two related recognition patterns, $f'(x)\big[f(x)\big]^n$ and $f'(x)e^{f(x)}$, which are the chain rule in reverse.

Where the logarithm comes from

Differentiate $\ln f(x)$ with the chain rule. The outer function is $\ln$, whose derivative is $\dfrac{1}{\text{inside}}$, and the inside differentiates to $f'(x)$:

$$\frac{d}{dx}\ln f(x) = \frac{1}{f(x)}\cdot f'(x) = \frac{f'(x)}{f(x)}$$

Read backwards, that is an integral. As with $\displaystyle\int\frac{1}{x}\,dx$, the modulus makes the result valid where $f(x)$ is negative.

So the test is: differentiate the denominator and compare with the numerator. If they match up to a constant factor, the answer is that constant times $\ln\lvert\text{denominator}\rvert$.

Integrand	Derivative of denominator	$k$	Integral
$\dfrac{2x}{x^2 + 5}$	$2x$	$1$	$\ln(x^2 + 5) + c$
$\dfrac{x}{x^2 + 5}$	$2x$	$\tfrac{1}{2}$	$\tfrac{1}{2}\ln(x^2 + 5) + c$
$\dfrac{x^2}{x^3 - 1}$	$3x^2$	$\tfrac{1}{3}$	$\tfrac{1}{3}\ln\lvert x^3 - 1\rvert + c$
$\dfrac{e^{2x}}{e^{2x} + 3}$	$2e^{2x}$	$\tfrac{1}{2}$	$\tfrac{1}{2}\ln(e^{2x} + 3) + c$
$\dfrac{\cos x}{2 + \sin x}$	$\cos x$	$1$	$\ln(2 + \sin x) + c$
$\dfrac{1}{x\ln x}$	$\dfrac{1}{x}$	$1$	$\ln\lvert\ln x\rvert + c$

The modulus can be dropped when the expression is always positive, as with $x^2 + 5$, $e^{2x} + 3$ and $2 + \sin x$. Keeping it is never wrong.

The constant $k$ can only fix a constant mismatch. $\dfrac{1}{x^2 + 5}$ does not fit the pattern, because the missing factor is $2x$, which depends on $x$. That integrand is an inverse tangent (see Standard integrals).

The integrals of tan x and cot x

The syllabus singles out $\tan x$. Write it as a quotient and look at the denominator:

$$\int \tan x\, dx = \int \frac{\sin x}{\cos x}\, dx = -\int \frac{-\sin x}{\cos x}\, dx = -\ln\lvert\cos x\rvert + c$$

Because $-\ln\lvert\cos x\rvert = \ln\lvert\cos x\rvert^{-1}$, this is also $\ln\lvert\sec x\rvert + c$. Similarly the numerator of $\cot x = \dfrac{\cos x}{\sin x}$ is exactly the derivative of the denominator.

Splitting the numerator

Often the numerator is the derivative of the denominator plus something else. Split the fraction into a part that fits the pattern and a part that does not, and deal with each separately.

$$\frac{x + 3}{x^2 + 9} = \frac{x}{x^2 + 9} + \frac{3}{x^2 + 9}$$

The first part is $\tfrac{1}{2}\cdot\dfrac{2x}{x^2 + 9}$, a logarithm. The second is an inverse tangent with $a = 3$:

$$\int \frac{x + 3}{x^2 + 9}\, dx = \tfrac{1}{2}\ln(x^2 + 9) + \tan^{-1}\left(\frac{x}{3}\right) + c$$

This exact split is what you need when a partial fraction has a quadratic denominator $x^2 + a^2$; see Integrating with partial fractions.

An improper fraction, where the numerator's degree is at least the denominator's, should be divided first: $\dfrac{2x + 5}{x + 1} = 2 + \dfrac{3}{x + 1}$, which integrates to $2x + 3\ln\lvert x + 1\rvert + c$.

Related patterns: the chain rule in reverse

The same idea, "spot the inside and its derivative", works for two more shapes. These are not separate syllabus statements, but recognising them saves time; anything they do can also be done with a substitution.

Each is the chain rule read backwards: $\dfrac{d}{dx}\big[f(x)\big]^{n+1} = (n + 1)\big[f(x)\big]^n f'(x)$ and $\dfrac{d}{dx}e^{f(x)} = f'(x)e^{f(x)}$. As before, a constant factor mismatch is fine; an $x$-dependent one is not.

• $\displaystyle\int x(x^2 + 1)^4\, dx = \tfrac{1}{2}\int 2x(x^2 + 1)^4\, dx = \tfrac{1}{10}(x^2 + 1)^5 + c$.
• $\displaystyle\int \cos x\sin^3 x\, dx = \tfrac{1}{4}\sin^4 x + c$, since $\cos x$ is the derivative of $\sin x$.
• $\displaystyle\int \sec^2 x\tan^2 x\, dx = \tfrac{1}{3}\tan^3 x + c$.
• $\displaystyle\int x e^{x^2}\, dx = \tfrac{1}{2}e^{x^2} + c$.

Odd powers of sine and cosine

Odd powers can be put into the power pattern with $\sin^2 x + \cos^2 x = 1$. Keep one factor aside as the derivative:

$$\cos^3 x = \cos x\cos^2 x = \cos x\left(1 - \sin^2 x\right) = \cos x - \cos x\sin^2 x$$ $$\int \cos^3 x\, dx = \sin x - \tfrac{1}{3}\sin^3 x + c$$

Even powers need the double angle identities instead (see Integrating using trigonometric identities).

Worked examples

The shaded area under $y = \dfrac{x + 2}{x^2 + 4}$ from $0$ to $2$ is $\tfrac{1}{2}\ln 2 + \tfrac{\pi}{4} \approx 1.132$: a logarithm and an inverse tangent from one fraction.

Practice

Answers

1. The numerator is exactly the derivative of $x^3 + 4$: $\ln\lvert x^3 + 4\rvert + c$.

2. $\dfrac{d}{dx}(2x^2 + 3) = 4x$, and $x = \tfrac{1}{4}(4x)$: $\tfrac{1}{4}\ln(2x^2 + 3) + c$.

3. $\Big[\ln\left(e^x + 2\right)\Big]_0^1 = \ln(e + 2) - \ln 3 = \ln\dfrac{e + 2}{3}$.

4. $\Big[\ln(\sin x)\Big]_{\pi/6}^{\pi/2} = \ln 1 - \ln\tfrac{1}{2} = \ln 2$.

5. $\sec^2 x$ is the derivative of $1 + \tan x$: $\Big[\ln(1 + \tan x)\Big]_0^{\pi/4} = \ln 2 - \ln 1 = \ln 2$.

6. Power pattern with $f(x) = x^2 + 1$, $f'(x) = 2x$: $\displaystyle\int x(x^2 + 1)^3\, dx = \tfrac{1}{8}(x^2 + 1)^4 + c$. Between the limits: $\tfrac{1}{8}(16 - 1) = \dfrac{15}{8}$.

7. $\cos^3 x = \cos x - \cos x\sin^2 x$, so the integral is $\Big[\sin x - \tfrac{1}{3}\sin^3 x\Big]_0^{\pi/2} = 1 - \tfrac{1}{3} = \dfrac{2}{3}$.

8. Split: $\dfrac{2x}{x^2 + 3} - \dfrac{1}{x^2 + 3}$. Integral: $\Big[\ln(x^2 + 3) - \tfrac{1}{\sqrt{3}}\tan^{-1}\tfrac{x}{\sqrt{3}}\Big]_0^{\sqrt{3}} = (\ln 6 - \ln 3) - \tfrac{1}{\sqrt{3}}\cdot\tfrac{\pi}{4} = \ln 2 - \dfrac{\pi}{4\sqrt{3}}$ (or $\ln 2 - \dfrac{\sqrt{3}\,\pi}{12}$).

9. The derivative of $\cos x + \sin x$ is $\cos x - \sin x$, the numerator. So the integral is $\Big[\ln(\cos x + \sin x)\Big]_0^{\pi/4} = \ln\sqrt{2} - \ln 1 = \tfrac{1}{2}\ln 2$. Over $[0, \pi]$ the denominator $\cos x + \sin x = \sqrt{2}\sin\left(x + \tfrac{\pi}{4}\right)$ is zero at $x = \tfrac{3\pi}{4}$, so the integrand is undefined there and the integral does not exist as an ordinary definite integral.

10. (a) Multiply out: $\sec x(\sec x + \tan x) = \sec^2 x + \sec x\tan x$, so the right-hand side equals $\dfrac{\sec x(\sec x + \tan x)}{\sec x + \tan x} = \sec x$. (b) $\dfrac{d}{dx}(\sec x + \tan x) = \sec x\tan x + \sec^2 x$, which is the numerator. So $\displaystyle\int\sec x\, dx = \ln\lvert\sec x + \tan x\rvert + c$. At $\tfrac{\pi}{4}$: $\sec\tfrac{\pi}{4} + \tan\tfrac{\pi}{4} = \sqrt{2} + 1$. At $0$: $1 + 0 = 1$. The value is $\ln(1 + \sqrt{2}) - \ln 1 = \ln\left(1 + \sqrt{2}\right)$.