Add one to the power, divide, then + C: Integration undoes differentiation. For a power: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C (for n ≠ −1) — add 1 to the power, divide by the new power, and always add the constant of integration C.
IB-style question — a single power
Find ∫x³ dx.
Step by step
- Add 1 to the power, divide by the new power.
- Add the constant.
Final answer
x⁴/4 + C.
Never forget + C: An indefinite integral always needs + C — leaving it off loses a mark.
Integrate term by term, one + C at the end: Integrate each term with the power rule, keep the signs, and add a single + C for the whole expression.
IB-style question — a polynomial
Find ∫(6x² − 4x + 5) dx.
Step by step
- Integrate each term.
- Simplify and add C.
Final answer
2x³ − 2x² + 5x + C.
Constant → ×x: A constant term like 5 integrates to 5x (think 5 = 5x⁰, then add 1 to the power).
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Rewrite, then add 1 to the power: Rewrite fractions and roots as powers first. ∫x⁻² dx = x⁻¹/(−1) + C = −1/x + C; ∫√x dx = ∫x^(1/2) dx = (2/3)x^(3/2) + C. (The rule excludes n = −1.)
IB-style question — root and fraction
Find ∫√x dx and ∫(1/x²) dx.
Step by step
- √x = x1/2: add 1 → 3/2, divide.
- 1/x² = x⁻²: add 1 → −1, divide.
Final answer
(2/3)x3/2 + C and −1/x + C.
Dividing by a fraction: Dividing by 3/2 means multiplying by 2/3 — a common slip when integrating roots.
Integrate, then use a point to find C: Given f'(x) and a point on the curve, integrate to get f(x) with a + C, then substitute the point to solve for C.
IB-style question — find f
A curve has f'(x) = 3x² − 2 and passes through (1, 4). Find f(x).
Step by step
- Integrate f'.
- Use f(1) = 4 to find C.
Final answer
f(x) = x³ − 2x + 5.
The point pins down C: Without a point, you can only find f up to + C; the given point fixes the exact curve.