Choosing the Substitution
You've seen how substitution works — now the real skill is picking the right $u$. A well-chosen substitution turns a hard integral into a one-liner; a poorly-chosen one creates a dead end. This lesson builds your pattern-recognition: look for a composite function and its derivative lurking inside the integrand.
Consider $\displaystyle\int 2x\sqrt{x^2+1}\,dx$. Without calculating — what expression would you choose for $u$? Write your reasoning below.
Every integration-by-substitution problem has two ingredients hiding in the integrand: a function $f(x)$ and a scalar multiple of its derivative $f'(x)$. Your job is to spot them.
The chain-rule reversal pattern:
If integrand contains $f'(x)\cdot g(f(x))$, set $u = f(x)$, so $du = f'(x)\,dx$.
The $f'(x)\,dx$ part absorbs into $du$ and the integral becomes $\displaystyle\int g(u)\,du$ — often trivial.
Key facts
- The pattern $\int f'(x)[f(x)]^n\,dx = \dfrac{[f(x)]^{n+1}}{n+1} + C$
- When to use the inner function as $u$ vs. a less obvious substitution
- How to handle a missing constant factor in $du$
Concepts
- Why the chain rule underpins every substitution
- How "spotting $f'(x)$ beside $f(x)$" reduces guesswork
- The role of changing limits for definite integrals after substitution
Skills
- Identify a suitable substitution for integrals involving radicals and compositions
- Verify a chosen substitution by computing $du$ and checking it appears in the integrand
- Complete the substitution, integrate, and back-substitute correctly
The substitution rule works whenever the integrand can be written in the form $f'(x) \cdot g(f(x))$. Setting $u = f(x)$ gives $du = f'(x)\,dx$, so:
The most common application is the power pattern:
Worked application: $\displaystyle\int 2x\sqrt{x^2+1}\,dx$
Set $u = x^2+1$, so $du = 2x\,dx$. The factor $2x\,dx$ is already present:
$= \displaystyle\int u^{1/2}\,du = \dfrac{2}{3}u^{3/2} + C = \dfrac{2}{3}(x^2+1)^{3/2} + C$
The substitution rule works whenever the integrand can be written in the form $f'(x) \cdot g(f(x))$. Setting $u = f(x)$ gives $du = f'(x)\,dx$, so:
Pause — copy the recognition pattern: identify inner function $f(x)$, check $f'(x)$ is present as a factor, then set $u=f(x)$ into your book.
Quick check: For $\displaystyle\int 3x^2(x^3+5)^4\,dx$, the best choice of $u$ is:
We just saw the recognition pattern: look for $f'(x)\cdot g(f(x))$ in the integrand — identifying the inner function $f(x)$ and confirming its derivative $f'(x)$ is present (up to a constant). That raises a question: when $f'(x)$ is present but multiplied by the wrong constant (e.g.\ $2x$ instead of $x$ inside $x^2$), how do you balance the integral without changing its value? This card answers it → multiply inside by the needed constant and outside by its reciprocal.
Sometimes $f'(x)$ is present but multiplied by the wrong constant. You can always balance this by introducing a compensating factor outside the integral.
Example: $\displaystyle\int x(x^2+3)^5\,dx$
Set $u = x^2+3$, so $du = 2x\,dx$. We only have $x\,dx$, not $2x\,dx$. Rewrite:
$x\,dx = \tfrac{1}{2}\,du$
The $\tfrac{1}{2}$ factor is introduced outside the integral to compensate for the missing 2 in $du$.
Sometimes $f'(x)$ is present but multiplied by the wrong constant. You can always balance this by introducing a compensating factor outside the integral.
Pause — copy the constant-adjustment technique: $\int 2x\cdot f(x^2)\,dx$ vs $\int x\cdot f(x^2)\,dx$ — introduce $\frac{1}{2}\cdot 2$ to match the derivative into your book.
Did you get this? True or false: for $\displaystyle\int x^2(x^3-1)^7\,dx$, setting $u = x^3-1$ works because $du = 3x^2\,dx$ and $x^2\,dx = \tfrac{1}{3}du$.
Worked examples · 3 in a row, reveal as you go
Find $\displaystyle\int 6x^2(2x^3-1)^4\,dx$.
Find $\displaystyle\int \frac{x}{\sqrt{x^2+4}}\,dx$.
Evaluate $\displaystyle\int_0^2 x(x^2+1)^3\,dx$.
Fill the gap: For $\displaystyle\int x(x^2-3)^2\,dx$, setting $u = x^2-3$ gives $du = 2x\,dx$, so the integral becomes $\dfrac{1}{2}\displaystyle\int u^2\,du = \dfrac{u^3}{$$} + C$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: when evaluating $\displaystyle\int_1^3 2x(x^2+1)^3\,dx$ via $u = x^2+1$, the new limits are $u = 2$ and $u = 10$.
Activities · practice with the ideas
State a suitable substitution $u$ for $\displaystyle\int \cos x \cdot \sin^3 x\,dx$ and explain why it works.
Find $\displaystyle\int 4x(x^2-7)^3\,dx$.
Find $\displaystyle\int \frac{x^2}{\sqrt{x^3+1}}\,dx$.
Evaluate $\displaystyle\int_0^1 3x^2(x^3+2)^4\,dx$. Change the limits.
Explain why $u = x$ is never a useful substitution. What does a useful $u$ always achieve?
Odd one out: Three of these substitutions will work for their respective integrals. Which one will NOT?
Earlier you chose a substitution for $\displaystyle\int 2x\sqrt{x^2+1}\,dx$.
The correct choice is $u = x^2+1$, because its derivative $2x$ is the coefficient of the radical in the integrand. This gives $\displaystyle\int \sqrt{u}\,du = \tfrac{2}{3}u^{3/2} + C = \tfrac{2}{3}(x^2+1)^{3/2} + C$. Did you spot the pattern?
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. State a suitable substitution for $\displaystyle\int 5x^4(x^5+3)^2\,dx$ and find the integral. (1 mark)
Q2. Find $\displaystyle\int \frac{x}{\sqrt{x^2-9}}\,dx$, $x > 3$. Show your substitution and adjustment. (2 marks)
Q3. Evaluate $\displaystyle\int_0^{\pi/2} \sin^3 x \cos x\,dx$. Change the limits using your substitution. (2 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $u = \sin x$, $du = \cos x\,dx$ — perfect match. 2. $u = x^2-7$, $du = 2x\,dx$; $\int 4x(x^2-7)^3\,dx = 2\int u^3\,du = \frac{u^4}{2}+C = \frac{(x^2-7)^4}{2}+C$. 3. $u = x^3+1$, $du = 3x^2\,dx$; $x^2\,dx = \frac{1}{3}du$; $\frac{1}{3}\int u^{-1/2}\,du = \frac{2}{3}u^{1/2}+C = \frac{2}{3}\sqrt{x^3+1}+C$. 4. $u = x^3+2$; limits $u=2$ to $u=3$; $\left[\frac{u^5}{5}\right]_2^3 = \frac{243-32}{5} = \frac{211}{5}$. 5. $u = x$ gives $du = dx$ so nothing simplifies; a useful $u$ must simplify the integrand by replacing a composite expression.
Q1 (1 mark): $u = x^5+3$, $du = 5x^4\,dx$; $\int u^2\,du = \dfrac{u^3}{3}+C = \dfrac{(x^5+3)^3}{3}+C$ [1].
Q2 (2 marks): $u = x^2-9$, $du = 2x\,dx$, $x\,dx = \frac{1}{2}du$ [1]; $\frac{1}{2}\int u^{-1/2}\,du = u^{1/2}+C = \sqrt{x^2-9}+C$ [1].
Q3 (2 marks): $u = \sin x$, $du = \cos x\,dx$; limits: $x=0\Rightarrow u=0$, $x=\frac{\pi}{2}\Rightarrow u=1$ [1]; $\int_0^1 u^3\,du = \left[\frac{u^4}{4}\right]_0^1 = \frac{1}{4}$ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering substitution questions. A lighter alternative to the boss battle.
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