When the numerator of an integrand is the derivative of the denominator, the answer is a natural log, full stop. Recognising this pattern turns ugly-looking integrals into one-line answers. This lesson nails the $f'(x)/f(x)$ rule, then shows how integration by parts unlocks integrals like $\int \ln x\, dx$ that no substitution can crack.
+5 XP warm-up Differentiate using the chain rule, and read off what must equal.
Differentiate $\ln|f(x)|$ using the chain rule, and read off what $\int \dfrac{f'(x)}{f(x)}\, dx$ must equal. Before checking would the same logic work for $\int \dfrac{x}{x^2 + 1}\, dx$? What constant would you need to factor out?
+5 XP to read Logarithmic integrals reward two habits: scan the integrand for a derivative-over-function structure (numerator $=$ derivative of denominator, up to a constant), then when $\ln$ sits inside the integrand, reach for integration by parts with $u = \ln x$.
Logarithmic integrals reward two habits: scan the integrand for a derivative-over-function structure (numerator $=$ derivative of denominator, up to a constant), then when $\ln$ sits inside the integrand, reach for integration by parts with $u = \ln x$. Confusing these two roles, log as output vs log as input is the biggest cause of error.
The scan-classify-technique reading: (1) check if numerator is (a constant multiple of) the derivative of the denominator, (2) if yes use the log rule, (3) if no but $\ln$ appears, try integration by parts with $u = \ln x$.
Log rule: $\int \frac{f'(x)}{f(x)}\, dx = \ln|f(x)| + C$ · By parts: $\int u\, dv = uv - \int v\, du$
Differentiating $\ln|f(x)|$ by the chain rule gives $\frac{f'(x)}{f(x)}$.
Differentiating $\ln|f(x)|$ by the chain rule gives $\frac{f'(x)}{f(x)}$. Reversing this is the log rule for integration:
Worked through the hook: For $\int \dfrac{2x}{x^2 + 1}\, dx$, the denominator is $f(x) = x^2 + 1$, and the numerator $2x$ is exactly $f'(x)$. So the answer is $\ln(x^2 + 1) + C$ (no need for $|\cdot|$ since $x^2+1>0$).
If the numerator is off by a constant, factor it out. For $\int \dfrac{x}{x^2 + 1}\, dx$:
Log rule: $\int \frac{f'(x)}{f(x)}\, dx = \ln|f(x)| + C$ · If the numerator is off by a constant $k$, factor $\frac{1}{k}$ out the front · $\int \frac{1}{x}\, dx = \ln|x| + C$, keep the absolute value · Worked: $\int \frac{x}{x^2+1}\, dx = \tfrac{1}{2}\ln(x^2+1) + C$
Pause, copy the log rule $\int f'(x)/f(x)\,dx = \ln|f(x)|+C$, the constant-factor adjustment, and the worked example $\int x/(x^2+1)\,dx = \tfrac{1}{2}\ln(x^2+1)+C$ into your book.
Quick check: Which of the following equals $\displaystyle \int \frac{3x^2}{x^3 + 5}\, dx$?
We just saw the $f'(x)/f(x)$ log rule: adjust the numerator by a constant factor, then integrate to $\ln|f(x)|+C$; the standard result $\int\ln x\,dx$ is proved by IBP with $u = \ln x$.
We just saw the $f'(x)/f(x)$ log rule: adjust the numerator by a constant factor, then integrate to $\ln|f(x)|+C$; the standard result $\int\ln x\,dx$ is proved by IBP with $u = \ln x$. That raises a question: how does LIATE specifically handle integrands involving $\ln x$ multiplied by a power of $x$? This card answers it → always take $u = \ln(\ldots)$ (highest LIATE class), $dv = x^n dx$; this gives $\int x\ln x\,dx = \tfrac{x^2}{2}\ln x - \tfrac{x^2}{4}+C$.
When $\ln$ appears inside the integrand (not as the answer), substitution rarely helps. The product rule run in reverse, integration by parts, is the right tool:
Why $u = \ln x$? Differentiating $\ln x$ gives $\frac{1}{x}$, a much simpler function. So the new integral $\int v\, du$ has the $\ln$ gone, replaced by a power of $x$.
Standard derivation of $\int \ln x\, dx$:
By parts: $\int u\, dv = uv - \int v\, du$ · LIATE: when $\ln$ is present, choose $u = \ln(\ldots)$ · $\int \ln x\, dx = x \ln x - x + C$ (memorise) · $\int x \ln x\, dx = \tfrac{x^2}{2}\ln x - \tfrac{x^2}{4} + C$
Pause, copy the IBP rule for log integrals (choose $u = \ln(\ldots)$), $\int\ln x\,dx = x\ln x - x+C$ (memorise), and $\int x\ln x\,dx = \tfrac{x^2}{2}\ln x - \tfrac{x^2}{4}+C$ into your book.
Did you get this? True or false: $\displaystyle \int \ln x\, dx = \frac{1}{x} + C$.
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Evaluate $\displaystyle \int \frac{\cos x}{2 + \sin x}\, dx$.
Find $\displaystyle \int x^2 \ln x\, dx$.
Evaluate $\displaystyle \int \frac{(\ln x)^3}{x}\, dx$.
Fill the gap: Using integration by parts with $u = \ln x$ and $dv = dx$, we get $du = $ $\, dx$ and $v = $ , giving $\int \ln x\, dx = x \ln x - x + C$.
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\displaystyle \int \frac{x}{x^2 + 1}\, dx = \ln(x^2 + 1) + C$.
Activities · practice with the ideas
Activities · practice with the ideas
Evaluate $\displaystyle \int \frac{4x^3}{x^4 + 7}\, dx$.
Evaluate $\displaystyle \int \frac{x^2}{x^3 - 1}\, dx$.
Evaluate $\displaystyle \int \tan x\, dx$ by rewriting as $\int \frac{\sin x}{\cos x}\, dx$.
Find $\displaystyle \int \ln(2x)\, dx$ using integration by parts.
Evaluate $\displaystyle \int_1^e \frac{\ln x}{x}\, dx$ using the substitution $u = \ln x$.
Odd one out: Three of these integrals are direct applications of the $\int f'(x)/f(x)\, dx = \ln|f(x)| + C$ rule. Which one is NOT?
Earlier you tried $\int \dfrac{2x}{x^2 + 1}\, dx$ and $\int \ln x\, dx$, one a one-line log-rule, the other a by-parts derivation.
The first integrand was already in $f'(x)/f(x)$ form, so the answer was $\ln(x^2+1) + C$. The second has no $f'(x)/f(x)$ structure, no substitution alone can make the $\ln$ disappear. Integration by parts with $u = \ln x$ converts $\ln$ into $\frac{1}{x}$ after one differentiation, giving $\int \ln x\, dx = x \ln x - x + C$. Two different roles for $\ln$, answer vs ingredient, two different techniques.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Evaluate $\displaystyle \int \frac{6x^2 + 2}{x^3 + x}\, dx$. (2 marks)
Q2. Use integration by parts to evaluate $\displaystyle \int_1^e x \ln x\, dx$. (3 marks)
Q3. Find $\displaystyle \int \frac{1}{x \ln x}\, dx$ by using a substitution. State the substitution. (3 marks)
Activity answers:
1. $f(x) = x^4 + 7$, $f'(x) = 4x^3$. Integral $= \ln(x^4 + 7) + C$ (positive denominator, drop $|\cdot|$).
2. $f(x) = x^3 - 1$, $f'(x) = 3x^2$; numerator $x^2 = \tfrac{1}{3} \cdot 3x^2$. Integral $= \tfrac{1}{3} \ln|x^3 - 1| + C$.
3. $\int \tan x\, dx = \int \frac{\sin x}{\cos x}\, dx = -\int \frac{-\sin x}{\cos x}\, dx = -\ln|\cos x| + C = \ln|\sec x| + C$.
4. $u = \ln(2x)$, $dv = dx$ $\Rightarrow$ $du = \tfrac{1}{x}\, dx$, $v = x$. $\int \ln(2x)\, dx = x \ln(2x) - \int 1\, dx = x \ln(2x) - x + C$.
5. $u = \ln x$, $du = \tfrac{1}{x}\, dx$. Limits become $u = 0$ to $u = 1$. $\int_0^1 u\, du = \tfrac{1}{2}$.
Q1 (2 marks): $f(x) = x^3 + x$, $f'(x) = 3x^2 + 1$; numerator $6x^2 + 2 = 2(3x^2 + 1) = 2f'(x)$ [1]. Therefore $\int \frac{6x^2 + 2}{x^3 + x}\, dx = 2 \ln|x^3 + x| + C$ [1].
Q2 (3 marks): $u = \ln x$, $dv = x\, dx \Rightarrow du = \tfrac{1}{x}\, dx$, $v = \tfrac{x^2}{2}$ [1]. $\int_1^e x \ln x\, dx = \left[\tfrac{x^2}{2} \ln x\right]_1^e - \int_1^e \tfrac{x}{2}\, dx = \tfrac{e^2}{2} - \left[\tfrac{x^2}{4}\right]_1^e$ [1] $= \tfrac{e^2}{2} - \tfrac{e^2}{4} + \tfrac{1}{4} = \tfrac{e^2 + 1}{4}$ [1].
Q3 (3 marks): Substitution $u = \ln x$ so $du = \tfrac{1}{x}\, dx$ [1]. Then $\int \frac{1}{x \ln x}\, dx = \int \frac{1}{u}\, du$ [1] $= \ln|u| + C = \ln|\ln x| + C$ [1].
Sit the module quiz, then close the lesson off.
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