Integrate (ax + b) to a power, then divide by a: For a linear inside, integrate as if the bracket were x, then divide by the inner coefficient a: ∫(ax+b)ⁿ dx = (ax+b)ⁿ⁺¹ / [a(n+1)] + C.
(The ÷a undoes the chain rule's ×a.)
IB-style question — bracket power
Find ∫(2x + 1)⁴ dx.
Step by step
- Raise the power and divide by (new power × inner coefficient).
- Simplify and add C.
Final answer
(2x + 1)⁵/10 + C.
The ÷ a is essential: Forgetting to divide by a is the classic error — check by differentiating your answer.
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Same ÷ a for sin, cos, e and 1/(ax+b): ∫sin(ax+b)dx = −cos(ax+b)/a + C, ∫cos(ax+b)dx = sin(ax+b)/a + C, ∫eax+bdx = eax+b/a + C, and ∫1/(ax+b) dx = (1/a)ln|ax+b| + C.
Always divide by the inner coefficient a.
IB-style question — three at once
Find ∫sin(3x) dx, ∫e2x dx and ∫1/(2x + 1) dx.
Step by step
- Each integrates with a ÷ (inner coefficient).
- The reciprocal gives a log.
Final answer
−⅓cos(3x) + C; ½e2x + C; ½ln|2x + 1| + C.
1/(ax+b) → log: A reciprocal of a linear term integrates to a logarithm: (1/a)ln|ax+b|.
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Spot the inner derivative as a factor: If the integrand is (inner derivative) × (function of the inner), integrate the outer and keep the inner: e.g. ∫2x(x²+1)³ dx = (x²+1)⁴/4 + C.
The special case ∫ f'(x)/f(x) dx = ln|f(x)| + C (numerator is the derivative of the denominator).
IB-style question — recognise the pattern
Find ∫(2x)/(x² + 1) dx.
Step by step
- Numerator 2x is the derivative of the denominator x²+1.
- So it is the f'/f pattern.
Final answer
ln|x² + 1| + C.
Check: is the top the derivative of the bottom?: If yes, the integral is ln|bottom|.
If it's off by a constant factor, adjust by that constant.
IB-style question — simplify, then integrate
Find ∫ (3x² + 1) ⁄ x dx.
Step by step
- You can't integrate a quotient directly — split it into separate terms first.
- Now integrate term by term; ∫(1/x)dx = ln|x|.
Final answer
3⁄2 x² + ln|x| + C.