DP Math AA · HL · Calculus

AHL 5.16—Integration by substitution, parts and repeated parts

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Introduction to Integration by Substitution

Integration by substitution (sometimes called u-substitution) is the integration counterpart of the chain rule. It transforms a complicated integral into a simpler one by replacing a portion of the integrand with a new variable.

The core idea: if your integral has the structure

∫kf(g(x))g′(x)dx

you can let u=g(x), so that du=g′(x)dx. The integral then becomes:

k∫f(u)du

which , if you chose u wisely , is much easier to evaluate.

Integration by Substitution: A technique that simplifies an integral by replacing a composite expression g(x) with a new variable u, converting ∫f(g(x))g′(x)dx into ∫f(u)du.

Note

In IB examinations, if the integral is not already in the form ∫kg′(x)f(g(x))dx, the substitution will be provided for you in the question. When it is provided, your job is to apply it correctly , not to invent it.

The method only helps if the resulting integral in u is simpler. Always think before choosing your substitution: will this actually make things easier?

Steps for Integration by Substitution

Follow these five steps systematically:

  1. Identify a suitable substitution u=g(x) , look for an inner function whose derivative also appears in the integrand.
  2. Differentiate to find dxdu​=g′(x), then rearrange to express dx (or g′(x)dx) in terms of du.
  3. Rewrite the entire integral in terms of u only , no x should remain.
  4. Integrate with respect to u.
  5. Back-substitute to express the answer in terms of x.
Exam Tip

Look for a function and its derivative both present in the integrand , that pairing is almost always the right choice for u. For example, in ∫xex2dx, notice that x is (up to a constant) the derivative of x2.

Warning

For definite integrals, when you substitute u=g(x) you must also change the limits. The lower limit x=a becomes u=g(a) and the upper limit x=b becomes u=g(b). Failing to change the limits is one of the most common errors in exam scripts.

Example

Example: Evaluate ∫x1−x2​dx

Step 1: Let u=1−x2

Step 2: dxdu​=−2x⟹du=−2xdx⟹xdx=−21​du

Step 3: The integral becomes:
−21​∫u​du=−21​∫u1/2du

Step 4: Integrate:
−21​⋅32​u3/2+C=−31​u3/2+C

Step 5: Back-substitute u=1−x2:
−31​(1−x2)3/2+C​

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10 more sections in this topic

← Previous topicAHL 5.15—Further derivatives and indefinite integration of these, partial fractionsNext topic →AHL 5.17—Areas under curve onto y-axis, volume of revolution (about x and y axes)
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