To integrate a power of x, add one to the power and divide by the new power: ∫xⁿ dx = xn+1/(n + 1) + c, for n ≠ −1. Constant multipliers carry through unchanged.
This is the working skill inside every integration question. It follows recovering a function from its derivative, and it needs the index work from earlier Additional Mathematics.
How do you handle awkward powers?
Rewrite every term as a number times a power of x before you integrate. Then the same rule applies every time.
| Original | As a power | Integral |
|---|---|---|
| x⁴ | x⁴ | x⁵/5 |
| 1/x³ | x−3 | x−2/(−2) = −1/(2x²) |
| √x | x1/2 | x3/2/(3/2) = (2/3)x3/2 |
| 1/√x | x−1/2 | x1/2/(1/2) = 2√x |
Dividing by a fraction means multiplying by its reciprocal. That is why dividing by 3/2 gives a factor of 2/3.
How do you integrate a full expression, step by step?
- Expand or split so that each term is a constant times a single power.
- Integrate each term with the rule, keeping any coefficient.
- Simplify the coefficients carefully, especially with negatives and fractions.
- Add + c once at the end.
Worked example
Find ∫(6x² − 4/x³ + 3√x) dx.
Step 1, rewrite: 6x² − 4x−3 + 3x1/2.
Step 2, integrate each term:
- 6x² gives 6 × x³/3 = 2x³.
- −4x−3 gives −4 × x−2/(−2) = 2x−2.
- 3x1/2 gives 3 × x3/2/(3/2) = 3 × (2/3)x3/2 = 2x3/2.
Step 3, combine:
2x³ + 2/x² + 2x3/2 + c
Step 4, check: differentiate the answer. You get 6x² − 4x−3 + 3x1/2, which is the original expression.
The mistake to watch for
A common slip is to lose the negative sign when the new power is negative.
Mistaken answer: ∫4/x³ dx = 4x−2/2 = 2x−2 + c
The new power is −2, so the divisor is −2, not 2. The sign was dropped.
The correct answer is 4x−2/(−2) = −2x−2 + c. Write the divisor with its sign in brackets, such as ”÷ (−2)”, before you simplify. Differentiating −2x−2 gives 4x−3, which confirms it.
Check yourself
Try these on paper, then open each answer.
1. Find ∫(5x⁴ − 2x) dx.
Show answer
5x⁴ gives x⁵, and −2x gives −x². x⁵ − x² + c
2. Find ∫(1/√x) dx.
Show answer
1/√x = x−1/2. Adding one gives x1/2, and dividing by 1/2 gives 2x1/2. 2√x + c
3. Find ∫(x + 1)² dx by expanding first.
Show answer
(x + 1)² = x² + 2x + 1. Integrating gives x³/3 + x² + x. x³/3 + x² + x + c
Where this leads next
Next, extend the idea to brackets in using a substitution-shaped reverse derivative. Always finish by checking an antiderivative by differentiation. The calculus shape and rate explorer and the non-calculator working trainer are useful for practice.
If fractions and negative indices keep slowing you down, that is a pattern our teachers can work on in online one-to-one Additional Mathematics tuition.