1The idea in one paragraph

To integrate a function f is to find a function F whose derivative is f. We write ∫ f(x) dx = F(x) + C and call F an antiderivative of f. The constant of integration C is there because a constant differentiates to zero, so F(x) + 5 and F(x) − 2 have the same derivative; Figure 1 shows the whole family. Every rule on this page is a differentiation rule read backwards, and every answer can be checked by differentiating it. The standard integrals cover powers, sine, cosine, eˣ and 1/x. The chain rule, read backwards, handles two more cases: a linear inside, ax + b, where you divide by a; and any inside g(x) whose derivative g′(x) is already sitting in the integrand, where you integrate the outside and keep the inside. That second case is integration by inspection, also called the reverse chain rule, and integration by substitution is the same idea written out in full.

2The standard integrals

Each line below is a derivative you already know, turned round.

A constant multiple comes along unchanged, and a sum integrates term by term: ∫(3 sin x + 4eˣ) dx = −3 cos x + 4eˣ + C.

Add 1 to the power, divide by the new power. And check every answer by differentiating it back: if you do not get the integrand, the answer is wrong.

The sine and cosine signs are the classic slip. Figure 2 puts the four functions on a cycle. Differentiating goes clockwise: sin → cos → −sin → −cos → sin. Integrating goes the other way. So ∫ sin x dx lands on −cos x, and ∫ cos x dx lands on sin x. All of this is in radians.

Rewrite before you integrate. The power rule only works on a sum of terms of the form axⁿ, so turn everything into that shape first. Roots become fractional powers, reciprocals become negative powers, brackets get expanded, and a fraction with a single term on the bottom gets split.

Dividing by a fraction is where marks go: x^3/2 ÷ (3/2) is (2/3)x^3/2. And the negative power flips a sign: −4 × x⁻¹ ÷ (−1) = +4x⁻¹.

There is no product rule and no quotient rule for integration. ∫(2x − 1)(x + 3) dx is not a product of two integrals; expand it to ∫(2x² + 5x − 3) dx = (2/3)x³ + (5/2)x² − 3x + C.

3Why ∫ 1/x dx is a logarithm

The power rule breaks at n = −1: it would give x⁰/0, and you cannot divide by zero. Yet 1/x is a perfectly good function, and it does have an antiderivative, because the derivative of ln x is 1/x. So ∫ 1/x dx = ln x + C, for x > 0.

What about negative x, where ln x does not exist? The function ln(−x) is defined there, and by the chain rule its derivative is (−1) × 1/(−x) = 1/x as well. Both cases together are ln|x|. Figure 3 shows the two curves side by side: at x = 2 the gradient of ln|x| is ½, and at x = −2 it is −½, exactly the heights of 1/x at those points.

∫ 1/x dx = ln|x| + C. When x > 0, which is most questions, this is ln x + C.

In practice, write ln|x| + C unless the question tells you x > 0; the modulus is never wrong.

4A linear inside: divide by a

Differentiate sin(2x + 3) and the chain rule multiplies by 2: the derivative is 2 cos(2x + 3). So to integrate cos(2x + 3), you need something whose derivative has no 2 in front, and that is ½ sin(2x + 3). This is the guide's own example:

The rule is general. If ∫ f(x) dx = F(x) + C, then

∫ f(ax + b) dx = (1/a) F(ax + b) + C. Integrate as if the bracket were x, then divide by a.

This only works because the derivative of ax + b is the constant a. With any other inside it fails. ∫(x² + 1)⁴ dx is not (x² + 1)⁵ ÷ (5 × 2x): differentiate that "answer" with the quotient rule and you will not get (x² + 1)⁴ back. That integral has to be expanded.

5The reverse chain rule: integration by inspection

The chain rule says d/dx [f(g(x))] = g′(x) f′(g(x)). Read backwards: if an integrand is the derivative of an inside function, multiplied by some function of that inside, then the integral is the outer function of the inside. In the guide's words, integrals of the form

∫ k g′(x) f(g(x)) dx = k F(g(x)) + C, where F is an antiderivative of f and k is a constant.

The skill is seeing it. Look for an inside function g(x), then look for its derivative g′(x) elsewhere in the integrand, allowing a constant factor. Figure 4 marks the pattern in the guide's three examples.

Example 1. ∫ 2x(x² + 1)⁴ dx. The inside is x² + 1 and its derivative 2x is right there. So the answer is "the integral of u⁴ with u = x² + 1":

Example 2. ∫ 4x sin(x²) dx. The inside is x², whose derivative is 2x; the integrand has 4x, which is 2 × 2x, so k = 2.

Example 3. ∫ sin x ÷ cos x dx. The inside is cos x, and its derivative is −sin x. The integrand has sin x, which is −1 times that, so k = −1, and the outer function is 1/u, whose integral is ln|u|.

Example 3 is one of a whole family worth recognising on sight: when the top is the derivative of the bottom,

∫ g′(x)/g(x) dx = ln|g(x)| + C.

So ∫ 3x²/(x³ + 1) dx = ln|x³ + 1| + C, and ∫ x²/(x³ + 1) dx = ⅓ ln|x³ + 1| + C, because x² is ⅓ of the derivative 3x².

The inspection method in three moves: guess the answer by integrating the outside and keeping the inside; differentiate your guess; adjust the constant so the derivative matches. You can only fix a constant factor this way. If ∫ x(x² + 1)⁴ dx had x² instead of x in front, no constant would fix it, and the method would not apply.

6Integration by substitution

Substitution is inspection with every step written down, and it is the safer method when the pattern is not obvious or the question says "use the substitution u = …".

1. Let u = g(x), the inside function.
2. Differentiate: du/dx = g′(x), so write du = g′(x) dx.
3. Replace every x and the dx, so the integral is in u alone.
4. Integrate with respect to u.
5. Substitute back u = g(x).

Example 4. ∫ x√(3x² + 1) dx.

Example 5. ∫ e^{sin x} cos x dx. With u = sin x, du = cos x dx, so the integral is ∫ eᵘ du = eᵘ + C = e^{sin x} + C.

Example 6. ∫ (ln x)/x dx. With u = ln x, du = (1/x) dx, so the integral is ∫ u du = u²/2 + C = (ln x)²/2 + C.

If, after substituting, an x is left that you cannot write in terms of u, stop: either the substitution is wrong or the integral is not of this form. ∫ e^x² dx is the standard example. There is no 2x outside, so u = x² leaves an x behind, and in fact no SL method finds this integral at all. Its definite values come from the GDC (5.11).

Figure 5 puts the whole page into four questions to ask of any integrand.

7Finding the constant

A boundary condition (a point the curve passes through, or a value at a given time) fixes C, exactly as in 5.5. Integrate first, then substitute, then write the final function in full.

Example 7. f′(x) = 6 sin 3x + 2 and f(0) = 5.

Note cos 0 = 1, not 0; the most common error in this question is to lose that −2.

Example 8. A curve has gradient dy/dx = 2x/(x² + 1) and passes through (0, 3).

This is the teal curve in Figure 1: of the whole family, it is the one through (0, 3).