Integration with Logarithms
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.
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?
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: $\int u\, dv = uv - \int v\, du$
Key facts
- $\int \frac{1}{x}\, dx = \ln|x| + C$ for $x \neq 0$
- $\int \frac{f'(x)}{f(x)}\, dx = \ln|f(x)| + C$
- Integration by parts: $\int u\, dv = uv - \int v\, du$
- $\int \ln x\, dx = x \ln x - x + C$
Concepts
- Why the log rule is just the chain rule run in reverse
- Why $u = \ln x$ is the right choice in by-parts (the $\ln$ disappears after one differentiation)
- Why a substitution alone cannot integrate $\ln x$
Skills
- Recognise $f'(x)/f(x)$ patterns and adjust by a constant
- Apply integration by parts with $u = \ln x$
- Combine substitution with the log rule for compound integrands
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 $\int \dfrac{x}{x^2 + 1}\, dx$:
- $f(x) = x^2 + 1$, $f'(x) = 2x$; the numerator is $x = \tfrac{1}{2}\cdot 2x$.
- So $\int \dfrac{x}{x^2 + 1}\, dx = \tfrac{1}{2} \int \dfrac{2x}{x^2 + 1}\, dx = \tfrac{1}{2}\ln(x^2 + 1) + C$.
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 $u = \ln x$. 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 $\int \ln x\, dx$:
- Let $u = \ln x$ and $dv = dx$. Then $du = \frac{1}{x}\, dx$ and $v = x$.
- $\int \ln x\, dx = x \ln x - \int x \cdot \tfrac{1}{x}\, dx = x \ln x - \int 1\, dx = x \ln x - x + C$.
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
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
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
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.
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)
Comprehensive answers (click to reveal)
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].
Five timed questions on the log rule, by-parts with $\ln x$, and combined substitution. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick integration questions. Lighter alternative to the boss.
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