Use a table of antiderivatives and u-substitution to integrate functions that hide a chain rule.
Why this matters
The fundamental theorem says that to integrate f, find a function whose derivative is f. For powers that is easy, but many integrals you meet in physics, chemistry and statistics hide a function inside another function. Substitution undoes the chain rule and handles most of them. You will rarely integrate complicated expressions by hand at work, since tools like SymPy, Wolfram Alpha and numerical integration do it, but recognising the patterns helps you set up models and check results.
A table of basic antiderivatives
| f(x) | ∫ f(x) dx |
|---|---|
| xⁿ (n ≠ −1) | xⁿ⁺¹ / (n + 1) + C |
| 1/x | ln|x| + C |
| eˣ | eˣ + C |
| e^(kx) | e^(kx) / k + C |
| sin x | −cos x + C |
| cos x | sin x + C |
Every entry can be checked by differentiating the right-hand side. That habit, differentiate your answer, catches almost every integration error.
The core idea of substitution
The chain rule says the derivative of F(g(x)) is F′(g(x)) · g′(x). Read backwards: if an integral has the form "a function of g(x), times g′(x)", substitute u = g(x) and du = g′(x) dx, and it collapses to a simple integral in u.
The steps:
- Choose u to be the "inside" expression, often the part inside parentheses, a root, or an exponent.
- Compute du = u′(x) dx.
- Rewrite the whole integral in terms of u. Nothing with x should remain.
- Integrate in u.
- Substitute back u = g(x).
Worked example 1
∫ 2x (x² + 1)⁵ dx
Let u = x² + 1. Then du = 2x dx, which is exactly the factor in front.
∫ u⁵ du = u⁶ / 6 + C = (x² + 1)⁶ / 6 + C
Check: the derivative of (x² + 1)⁶ / 6 is 6(x² + 1)⁵ · 2x / 6 = 2x(x² + 1)⁵. Correct.
Worked example 2: fixing a constant
∫ x · e^(3x²) dx
Let u = 3x², so du = 6x dx and x dx = du / 6.
∫ e^u · (du / 6) = e^u / 6 + C = e^(3x²) / 6 + C
Constant factors are easy to fix. If the leftover piece involved x itself (say x² instead of x), substitution with this u would not work and you would try a different u or a different method.
Definite integrals: change the limits
With limits, convert them to u-values as well and you never need to substitute back:
∫ from 0 to 1 of 2x (x² + 1)³ dx, u = x² + 1
x = 0 → u = 1, x = 1 → u = 2
∫ from 1 to 2 of u³ du = [u⁴ / 4] from 1 to 2 = 4 − 1/4 = 3.75
A midpoint sum in code gives the same 3.75, a quick way to confirm.
Other techniques, briefly
- Integration by parts reverses the product rule: ∫ u dv = uv − ∫ v du. It handles products like x·eˣ or x·ln x.
- Partial fractions split rational functions such as 1 / ((x − 1)(x + 2)) into simpler pieces; they appear in reaction kinetics and control systems.
- Numerical integration (trapezoid, Simpson's rule,
scipy.integrate.quad) is the workhorse when no closed form exists, for example the bell curve e^(−x²), whose antiderivative cannot be written with elementary functions. That is why normal-distribution probabilities come from tables or software.
Common mistakes
- Leaving some x behind. After substituting, the integral must be entirely in u and du.
- Forgetting du. Replacing g(x) with u but not dx with du / g′(x) gives the wrong constant.
- Keeping the old limits after switching to u in a definite integral.
- Skipping the check. Differentiating your answer takes seconds.
Quick recap
- Know the basic antiderivatives; check any answer by differentiating it.
- Substitution reverses the chain rule: pick the inside function as u, replace dx using du.
- In definite integrals, convert the limits to u.
- Integration by parts, partial fractions and numerical methods cover most of the rest.
Try it
This lesson has no code exercise. Work through the quiz in the Do tab, ideally with pencil and paper.