Integration by substitution
Substitution is the chain rule run backward. If you can spot a function and its derivative sitting in the same integral, you can swap the hard integral for an easy one.
1. See the chain rule hiding inside
Differentiating eg(x) gives eg(x)·g′(x). Read that backward and you have an antiderivative. Choose an inside function g to see it:
The giveaway: the derivative of the inside function appears as a factor. That’s the piece you’ll call u.
2. The method
- Choose u — usually the inside of a composition: under a root, in a power, in an exponent, in a denominator, inside sin/cos/ln.
- Differentiate: du = u′(x) dx. Solve for whatever piece of the integrand you need, like x dx = ½ du.
- Rewrite the entire integral in u. If any x remains, rethink your choice.
- Integrate in u, then replace u with its expression in x. Don’t forget + C.
Let u = 2x³+5. Then du = 6x² dx, which is exactly the factor in front. The integral becomes ∫ √u du = (2/3)u3/2 + C, so the answer is (2/3)(2x³+5)3/2 + C.
3. Definite integrals: change the bounds
The bounds belong to x. When you switch to u, convert them with u = g(x) and you never have to go back to x. Try it:
4. Practice
Work each problem one move at a time. Wrong choices give a hint and you can try again.
5. Patterns worth memorizing
| ∫ f′(x)/f(x) dx | = ln|f(x)| + C (let u = f(x)) |
| ∫ f′(x)·[f(x)]ⁿ dx | = [f(x)]ⁿ⁺¹/(n+1) + C |
| ∫ 1/(a²+x²) dx | = (1/a)·arctan(x/a) + C |
| ∫ 1/√(a²−x²) dx | = arcsin(x/a) + C |
6. Common mistakes
7. Quiz
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8. Free-response set
AP-style multi-part problems. Work each part on paper or in the box, open the solution, then mark yourself honestly. Marks are saved for the player selected in the quiz.