To learn Integration by Substitution quickly and correctly, look for an expression inside another expression, together with a factor related to its derivative. Substitution is a way of renaming that inner expression so a complicated-looking integral becomes a familiar one. It reverses the chain rule from differentiation. These eduKateSG Mathematics Quick Notes teach how to choose u, transform the entire integral, integrate, return to x and check your result by differentiating, with an extra lesson on how definite-integral limits must change.
- Look inside: identify a repeated bracket, denominator, exponent or trigonometric argument.
- Choose u: set u equal to a useful inner expression.
- Differentiate: find du/dx and express the relevant factor times dx in terms of du.
- Rewrite completely: the new integrand should use u, not a mixture of unrelated x and u terms.
- Integrate: use an ordinary antiderivative rule in u.
- Finish correctly: substitute back for an indefinite integral or convert the limits for a definite integral.
- Check: differentiate the result to recover the original integrand.
First, practise seeing the inside expression. Consider ∫2x(x² + 1)⁴ dx. The bracket x² + 1 is raised to a power, and its derivative is 2x, which appears as a multiplier in front. Let u = x² + 1. Then du = 2x dx. The original integral becomes ∫u⁴ du. You have not changed the value of the integral; you have written the same calculation using a simpler variable. This is why choosing a useful u can make a challenging expression look like a familiar power-rule question.
Second, integrate the simple expression and translate back. The antiderivative of u⁴ is u⁵/5 + C. Replace u with x² + 1, giving (x² + 1)⁵/5 + C. Differentiate it: (1/5) × 5(x² + 1)⁴ × 2x = 2x(x² + 1)⁴, exactly the original integrand. The derivative check is the best way to know you have not missed a factor. The constant C belongs in the final result because there were no numerical integration limits.
Third, learn that the derivative of your chosen u may not appear with exactly the same coefficient. For ∫6(3x + 1)⁴ dx, choose u = 3x + 1, so du = 3 dx. Since 6 dx = 2 du, the integral becomes 2∫u⁴ du = (2/5)u⁵ + C. Substitute back: (2/5)(3x + 1)⁵ + C. Check: derivative = (2/5) × 5(3x + 1)⁴ × 3 = 6(3x + 1)⁴. The method works because you adjusted the constant multiplier correctly.
Worked example: a bracket power with a variable factor
Integrate ∫4x(x² + 2)³ dx. Choose u = x² + 2, so du = 2x dx. Thus 4x dx = 2 du. Rewrite the integral as 2∫u³ du = 2(u⁴/4) + C = u⁴/2 + C. Therefore the answer is (x² + 2)⁴/2 + C. Check by differentiation: (1/2) × 4(x² + 2)³ × 2x = 4x(x² + 2)³. The exact match confirms the substitution and coefficient.
Worked example: a logarithmic antiderivative
Evaluate ∫[2x/(x² + 4)] dx. The denominator’s inner expression x² + 4 has derivative 2x, exactly the numerator. Set u = x² + 4 and du = 2x dx. The integral becomes ∫(1/u) du = ln|u| + C. Because x² + 4 is positive for every real x, the result can be written ln(x² + 4) + C. Differentiating gives 2x/(x² + 4), which matches the original integrand.
Worked example: a cosine with a linear inside
Find ∫cos(3x) dx in the usual radian-based calculus convention. Let u = 3x, so du = 3 dx and dx = du/3. Then the integral is (1/3)∫cos u du = (1/3)sin u + C. Replace u: sin(3x)/3 + C. Differentiate: (1/3) × 3cos(3x) = cos(3x). If you left out the factor 1/3, differentiating would give 3cos(3x), which is three times too large.
Worked example: exponential integration
Evaluate ∫2x e^(x²) dx. Use u = x², so du = 2x dx. The integral becomes ∫eᵘ du = eᵘ + C. Substitute back to obtain e^(x²) + C. Check with the chain rule: derivative of e^(x²) is 2x e^(x²). This is a particularly clear example of how substitution reverses differentiation: the derivative of the inside expression provides the multiplier needed outside.
Definite integrals: change the limits when u changes
Suppose you need ∫₀²2x(x² + 1)³ dx. Choose u = x² + 1, with du = 2x dx. Now change the numerical limits: when x = 0, u = 1; when x = 2, u = 5. The integral becomes ∫₁⁵u³ du. Evaluate u⁴/4 at the new limits: (5⁴ − 1⁴)/4 = (625 − 1)/4 = 156. There is no + C in the final definite value. You may alternatively integrate in u, return to x and then use the original x-limits, but do not mix u-limits with an x-expression.
A short worked definite logarithm integral
Find ∫₀¹[2x/(x² + 1)] dx. Set u = x² + 1, so du = 2x dx. The new limits are u = 1 when x = 0 and u = 2 when x = 1. Therefore the integral is ∫₁²(1/u) du = ln2 − ln1 = ln2. Since ln1 = 0, the final answer is ln2 ≈ 0.693. The integrand is non-negative on this interval, so a positive definite-integral result is sensible.
Why substitution is the reverse chain rule
When you differentiate a composition F(g(x)), the chain rule gives F′(g(x))g′(x). When you integrate an expression of this same form, you look for the inside g(x) and its derivative g′(x), then reverse the operation to recover F(g(x)) + C. For example, differentiating (x² + 1)⁵/5 gives 2x(x² + 1)⁴, so integrating that product returns the original expression. The structure is the central idea; the specific numbers are secondary.
How to choose a useful substitution
- Bracket raised to a power: try the expression inside the bracket.
- Function in the denominator: try the denominator when its derivative is present above.
- Exponential of a function: try the exponent.
- Sine or cosine of an expression: try the trigonometric argument.
- Constant multiple mismatch: adjust dx or the multiplier correctly.
- No useful inner derivative visible: substitution may not be the shortest method; investigate another integration rule.
A substitution does not always make an integral simple
Not every complicated integral becomes elementary merely because you rename one expression. For example, if you choose u = x² + 1 in ∫(x² + 1)⁴ dx, then du = 2x dx, but there is no x factor in the original integrand. You would still need to handle that mismatch, and the substitution may be unhelpful. For this polynomial example, expanding or another straightforward method is more suitable. The aim is a simpler, complete integral in the new variable, not simply replacing one bracket with a letter.
What if there is a square root in the denominator?
Consider ∫[3/√(3x + 2)] dx, on an interval where 3x + 2 > 0. Choose u = 3x + 2, giving du = 3 dx. The integral becomes ∫u^(−1/2) du = 2u^(1/2) + C. Substitute back: 2√(3x + 2) + C. Differentiate: 2 × [3/(2√(3x + 2))] = 3/√(3x + 2), as required. This also shows why keeping powers in fraction form can make the antiderivative easier to recognise.
The safest seven-step substitution method
- Identify the complete inner expression that would simplify the integrand.
- Write u = that expression.
- Differentiate to get du/dx and rearrange the relevant factor times dx.
- Replace every occurrence needed so the integral is fully in u.
- If the integral is definite, convert both limits to u-values now.
- Integrate in u; substitute back for an indefinite integral, or evaluate the converted limits for a definite one.
- Check by differentiating or by confirming the signed value and units.
Common substitution mistakes
- Changing u but leaving stray x terms in the integral: transform the entire integrand.
- Dropping a constant derivative: compensate for the factor when du differs from the numerator.
- Forgetting + C: indefinite antiderivatives need the arbitrary constant.
- Keeping x-limits after switching completely to u: convert definite limits to the new variable.
- Mixing u-limits with an answer rewritten in x: use one variable consistently at the evaluation stage.
- Applying the ordinary power rule to 1/u: integrate it as ln|u| where u ≠ 0.
- Choosing a substitution that simplifies nothing: look for an inner function with its derivative or a constant multiple.
- Skipping a derivative check: differentiate your final antiderivative to verify the original integrand.
Mini-test: solve the integral and check
- Find ∫4x(x² + 2)³ dx.
- Find ∫3(3x − 2)⁴ dx.
- Find ∫[2x/(x² + 1)] dx.
- Evaluate ∫₀¹[2x/(x² + 1)] dx.
- Evaluate ∫₀¹2x(x² + 1)² dx.
Answers with the substitution
1. u = x² + 2, du = 2x dx, giving (x² + 2)⁴/2 + C. 2. u = 3x − 2, du = 3 dx, giving (3x − 2)⁵/5 + C. 3. u = x² + 1 gives ln(x² + 1) + C. 4. The new limits are 1 and 2, so the answer is ln2. 5. With u = x² + 1, the new limits are 1 and 2; ∫₁²u² du = (2³ − 1³)/3 = 7/3.
Your twelve-minute substitution study routine
- Minutes 1–2: identify the inner expression in three candidate integrals.
- Minutes 3–5: integrate a bracket power after setting u.
- Minutes 6–7: integrate a rational expression that becomes 1/u.
- Minutes 8–10: solve a definite integral and explicitly convert its limits.
- Minutes 11–12: differentiate two final antiderivatives to check the results.
The key learning habit is to make the substitution complete. Choose an inside expression, account for its derivative, keep one variable at a time and verify the answer by reversing the calculation. That turns substitution from a memory trick into a reliable method.
Continue learning with connected Mathematics Quick Notes
Keep practising with Integration · Chain Rule Differentiation · Limits and Continuity · Logarithms. Solve one fresh problem without copying and explain the reason for every step.
The original teaching reference remains unchanged: Secondary 1 Mathematics Tutor Clementi | Small Groups Tutorials.