Mathematics 9709 · Pure Mathematics 3

Integrating to ln: when the top is the derivative of the bottom

Everything new in this chapter is just a differentiation rule run backwards.
To start a hard integral, ask what rule would have made this? Once you know that, the right method is clear.
This first one is the chain rule on a log, run backwards.

One idea runs the whole chapter. Before you integrate, run the rules backwards in your head. A fraction whose top is (a multiple of) the derivative of its bottom came from $\ln(\text{bottom})$ → $k\ln|f|$ . A product of two unrelated functions came from the product rule → by parts. A $\tfrac{1}{x^2+a^2}$ came from $\tan^{-1}$ → arctan. Spot the pattern first, integrate second. Picking the right method is where most of the marks are.

An integral is the area piling up under a curve. As the right edge slides along, the “area so far” traces out the antiderivative.
So integrating means finding the function whose growth rate is the curve you started with.
Every method below is just one way to spot which function was growing.

upper limit b

accumulated area A(b) = 5.63·which is exactly ½b² + b = 3.13 + 2.5

The right-hand curve is the antiderivative being drawn: its height at b is the area on the left. Integration runs this in reverse.

When the top is the derivative of the bottom

Differentiate $\ln|f(x)|$ with the chain rule and you get $\dfrac{f'(x)}{f(x)}$ : the derivative of the inside, over the inside.
Run that backwards. Any fraction shaped like $\dfrac{f'(x)}{f(x)}$ integrates straight back to $\ln|f(x)|$ .

$\displaystyle\int \frac{k\,f'(x)}{f(x)}\,dx = k\ln|f(x)| + c$ (textbook 8.3, Key Point 8.3)

The only test you run: differentiate the bottom.
If the top matches what you get (up to a constant), write $\ln$ of the bottom.
A quick test that works on fractions you could never factor.

The method: force the top to equal f′(x)

Look at the denominator $f(x)$ and differentiate it to get $f'(x)$ .
Check the numerator is $f'(x)$ up to a constant multiple.
Fix the constant: whatever you multiply the top by to make it exactly $f'(x)$ , divide that back out the front.
Write $k\ln|f(x)| + c$ — keep the modulus bars.

Worked example

Find $\int \frac{2x}{x^2+3}\,dx$

Bottom is $x^2+3$ . Differentiate it → $2x$ , which is exactly the top, so there is no constant to fix.

Answer

\int \frac{2x}{x^2+3}\,dx = \ln(x^2+3) + c

(textbook Worked Example 8.5). Here $x^2+3>0$ always, so the bars are optional. Write them anyway, out of habit.

Worked example

Find $\int \frac{3x^2+5x}{2x^3+5x^2}\,dx$ : fix the constant

Bottom $f=2x^3+5x^2$ ; differentiate → $f'=6x^2+10x$ .
The top is $3x^2+5x$ — that is exactly $\tfrac12 f'$ . So pull a $\tfrac12$ out front to make the top equal $f'$ .

Answer

\tfrac{1}{2}\ln\!\left|2x^3+5x^2\right| + c

Common mistake

Two easy marks get thrown away here. One is dropping the constant (forgetting the

\tfrac12

\tfrac13

out front when you force the top to be

f'

). The other is dropping the modulus bars. The answer must be

\ln|f(x)|

, not

\ln f(x)

Why you really do need the modulus bars

$\ln$ only takes a positive input, but a fraction like $\dfrac{1}{x}$ works fine on the negative side too. The modulus lets one formula cover both sides. For $x>0$ , $\tfrac{d}{dx}\ln x=\tfrac1x$ . For $x<0$ , $\ln|x|=\ln(-x)$ and the chain rule gives $\tfrac{1}{-x}\cdot(-1)=\tfrac1x$ again. So $\ln|x|$ differentiates to $\tfrac1x$ on both sides. That is why the antiderivative keeps the bars. Drop them and your answer is wrong wherever $f(x)<0$ .

The one to memorise: ∫ tan x dx = −ln|cos x|

This looks like it needs a special method, but it is just the $\tfrac{f'}{f}$ form. Write $\tan x = \dfrac{\sin x}{\cos x}$ .
The bottom is $\cos x$ . Differentiate it → $-\sin x$ .
The top is $+\sin x$ , so the top is $-1$ times $f'$ :

$\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx = -\ln|\cos x| + c \;(=\ln|\sec x|+c)$

Common mistake

The sign is the trap. Because

\cos x

differentiates to minus

\sin x

, the answer starts with a minus:

-\ln|\cos x|

, not

+\ln|\cos x|

. Same with

\tan 3x \to -\tfrac13\ln|\cos 3x|

. The

\tfrac13

is the chain-rule constant; the minus is the sign.

Where it shows up

This is the most useful thing to spot in the chapter. It shows up inside almost everything else: a piece of a partial-fraction integral (§8.4), the $\ln$ -half of a split arctan integral (§8.2), a term left over after a substitution.
Learn the test (“is the top the derivative of the bottom?”) and you can do all of them.

Your turn— tap to reveal the worked answer (9709/32 Nov 2023 Q5, the ln-part)

That paper integrates $\displaystyle\int_0^6 \frac{x(x+1)}{x^2+4}\,dx$ . After you divide out the improper top (§8.2), a $\dfrac{x}{x^2+4}$ piece is left over. The $x$ on top is half the derivative of $x^2+4$ (which is $2x$ ), so it is a $\tfrac{f'}{f}$ form:

Answer

\int \frac{x}{x^2+4}\,dx = \tfrac{1}{2}\ln(x^2+4) + c

The rest of that integral (the constant over $x^2+4$ ) is the arctan piece, and the divide-first step is §8.2. (9709/32 Nov 2023 Q5)

PreviousPartial fractions with binomial expansions NextThe arctan integral: 1/(x²+a²) and splitting (linear)/(x²+a²)

Integrating to ln: when the top is the derivative of the bottom

Everything new in this chapter is just a differentiation rule run backwards.
To start a hard integral, ask what rule would have made this? Once you know that, the right method is clear.
This first one is the chain rule on a log, run backwards.

An integral is the area piling up under a curve. As the right edge slides along, the “area so far” traces out the antiderivative.
So integrating means finding the function whose growth rate is the curve you started with.
Every method below is just one way to spot which function was growing.

upper limit b

accumulated area A(b) = 5.63·which is exactly ½b² + b = 3.13 + 2.5

The right-hand curve is the antiderivative being drawn: its height at b is the area on the left. Integration runs this in reverse.

When the top is the derivative of the bottom

Differentiate $\ln|f(x)|$ with the chain rule and you get $\dfrac{f'(x)}{f(x)}$ : the derivative of the inside, over the inside.
Run that backwards. Any fraction shaped like $\dfrac{f'(x)}{f(x)}$ integrates straight back to $\ln|f(x)|$ .

$\displaystyle\int \frac{k\,f'(x)}{f(x)}\,dx = k\ln|f(x)| + c$ (textbook 8.3, Key Point 8.3)

The only test you run: differentiate the bottom.
If the top matches what you get (up to a constant), write $\ln$ of the bottom.
A quick test that works on fractions you could never factor.

The method: force the top to equal f′(x)

Look at the denominator $f(x)$ and differentiate it to get $f'(x)$ .
Check the numerator is $f'(x)$ up to a constant multiple.
Fix the constant: whatever you multiply the top by to make it exactly $f'(x)$ , divide that back out the front.
Write $k\ln|f(x)| + c$ — keep the modulus bars.

Worked example

Find $\int \frac{2x}{x^2+3}\,dx$

Bottom is $x^2+3$ . Differentiate it → $2x$ , which is exactly the top, so there is no constant to fix.

Answer

\int \frac{2x}{x^2+3}\,dx = \ln(x^2+3) + c

(textbook Worked Example 8.5). Here $x^2+3>0$ always, so the bars are optional. Write them anyway, out of habit.

Worked example

Find $\int \frac{3x^2+5x}{2x^3+5x^2}\,dx$ : fix the constant

Bottom $f=2x^3+5x^2$ ; differentiate → $f'=6x^2+10x$ .
The top is $3x^2+5x$ — that is exactly $\tfrac12 f'$ . So pull a $\tfrac12$ out front to make the top equal $f'$ .

Answer

\tfrac{1}{2}\ln\!\left|2x^3+5x^2\right| + c

Common mistake

Two easy marks get thrown away here. One is dropping the constant (forgetting the

\tfrac12

\tfrac13

out front when you force the top to be

f'

). The other is dropping the modulus bars. The answer must be

\ln|f(x)|

, not

\ln f(x)

Why you really do need the modulus bars

The one to memorise: ∫ tan x dx = −ln|cos x|

This looks like it needs a special method, but it is just the $\tfrac{f'}{f}$ form. Write $\tan x = \dfrac{\sin x}{\cos x}$ .
The bottom is $\cos x$ . Differentiate it → $-\sin x$ .
The top is $+\sin x$ , so the top is $-1$ times $f'$ :

$\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx = -\ln|\cos x| + c \;(=\ln|\sec x|+c)$

Common mistake

The sign is the trap. Because

\cos x

differentiates to minus

\sin x

, the answer starts with a minus:

-\ln|\cos x|

, not

+\ln|\cos x|

. Same with

\tan 3x \to -\tfrac13\ln|\cos 3x|

. The

\tfrac13

is the chain-rule constant; the minus is the sign.

Where it shows up

This is the most useful thing to spot in the chapter. It shows up inside almost everything else: a piece of a partial-fraction integral (§8.4), the $\ln$ -half of a split arctan integral (§8.2), a term left over after a substitution.
Learn the test (“is the top the derivative of the bottom?”) and you can do all of them.

Your turn— tap to reveal the worked answer (9709/32 Nov 2023 Q5, the ln-part)

Answer

\int \frac{x}{x^2+4}\,dx = \tfrac{1}{2}\ln(x^2+4) + c

The rest of that integral (the constant over $x^2+4$ ) is the arctan piece, and the divide-first step is §8.2. (9709/32 Nov 2023 Q5)