Integration as anti-differentiation

Integration is differentiation run backwards. Differentiating x³ gives 3x², so integrating 3x² gives back x³. The answer is called an anti derivative (or primitive).

But x³ + 5 and x³ – 2 also have derivative 3x², because constants disappear when you differentiate. So we always add a constant of integration C: ∫3x² dx = x³ + C. An anti derivative is never unique.

Every standard integral is just a derivative you already know, read backwards:

• d/dx xⁿ⁺¹/(n + 1) = xⁿ, so ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C (n ≠ –1).
• d/dx log|x| = 1/x, so ∫(1/x) dx = log|x| + C.
• d/dx eˣ = eˣ, so ∫eˣ dx = eˣ + C.
• d/dx sin x = cos x, so ∫cos x dx = sin x + C.
• d/dx (–cos x) = sin x, so ∫sin x dx = –cos x + C.
• d/dx tan x = sec²x, so ∫sec²x dx = tan x + C.

Power rule in words: add 1 to the power, then divide by the new power. It fails only for n = –1 (dividing by 0), which is why 1/x has its own rule.

Integrals split over sums and let constants out: ∫(3x² + 4x – 5) dx = x³ + 2x² – 5x + C. ∫(2eˣ + 3/x) dx = 2eˣ + 3 log|x| + C.

If the expression is a fraction or product, simplify first: divide each term by the denominator, or expand the brackets, so that every term matches a standard form.

Finding C: if you know one value of F, put it in. F′(x) = 2x + 1 and F(1) = 4: F(x) = x² + x + C, so 1 + 1 + C = 4 and C = 2.