01 · Reverse a power
Find ∫8x³ dx.
Hint
Increase 3 to 4 and divide 8 by 4.
Worked solution
2x⁴ + C.
Understand · explore · practise
Integration: powers and polynomials
Integrate powers, polynomials, roots and reciprocals. Rewrite expressions, include the constant of integration and differentiate your answer to check it.
Before you startIndex laws, algebraic expansion and the differentiation power rule
01 / Reverse differentiation
Find a function with the gradient you were given.
Differentiating 2x³ gives 6x². Integration reverses that step: an antiderivative of 6x² is 2x³. But 2x³ + 7 and 2x³ − 4 have the same derivative too.
If F′(x) = f(x), then ∫f(x) dx = F(x) + C
The indefinite integral describes a family of functions. C is an arbitrary constant; it accounts for the information lost when constants differentiate to zero.
The most useful check is to differentiate your proposed answer. You should recover the complete original integrand.
02 / The integration power rule
Increase the power, then divide by the new power.
∫axⁿ dx = [a/(n + 1)]xn+1 + C, for n ≠ −1
The numbers a and n are constants. The rule holds on an interval where the powers and derivatives are defined as real functions. Increasing the power by one reverses differentiation; dividing by that new power reverses its multiplier.
∫4x³ dx = x⁴ + C
∫3x⁻² dx = −3x⁻¹ + C
∫5x1/2 dx = (10/3)x3/2 + C
Choose a coefficient and exponent. The model shows one antiderivative with C = 0, and checks its derivative at x = 1. The graphs use positive x-values so roots and reciprocals are defined.
Reverse a power ruleMove at your pace
Coefficient a = 3; exponent n = 2. An antiderivative with C = 0 is F(x) = x³. Its derivative at x = 1 is 3, equal to f(1). The graphs use x > 0. Blue is the integrand and green is the antiderivative; they are different functions.
Watch the exponent and coefficient reverse differentiation
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Constants and sums
A constant integrates to a linear term.
Because d(kx)/dx = k, the integral of a constant k is kx + C. Integrate a sum or difference term by term:
∫(6x² − 4x + 5) dx
= 2x³ − 2x² + 5x + C
The term 5 becomes 5x; it does not disappear. One arbitrary constant is enough for the whole expression because a sum of constants is another constant.
A zero function integrates to C. This is consistent with every horizontal line having derivative zero.
04 / Read the variable of integration
The differential tells you which letter can vary.
In ∫f(x) dx, f(x) is the integrand and dx names x as the variable of integration. Other letters are treated as constants unless a relationship says otherwise.
∫(px² + qx + 3) dx
= (p/3)x³ + (q/2)x² + 3x + C
∫(5t² − 2) dt = (5/3)t³ − 2t + C
Do not leave an integral sign on an evaluated answer. A formula using dt should produce an antiderivative in t, not silently switch to x.
05 / Rewrite roots and fractions
Expose a separate power in each term.
A fraction over a square rootWorked example
(3x² − 2x + 5)/√x
= 3x3/2 − 2x1/2 + 5x−1/2
Divide every term by x to the power 1/2; here x > 0.
∫(3x² − 2x + 5)/√x dx
= (6/5)x5/2 − (4/3)x3/2 + 10√x + C
Add one to each exponent, then divide by that new exponent.
Differentiate: 3x3/2 − 2x1/2 + 5x−1/2
Check all three coefficients and powers.
For a reciprocal, retain the original restriction x ≠ 0. More precisely, an antiderivative is chosen on an interval that does not cross zero; unrelated constants may be chosen on the two disconnected sides.
Why n = −1 is excluded
Adding one would give zero, so the displayed rule would divide by zero. The integral of 1/x requires a logarithmic rule from later calculus; it is not x⁰/0 or a constant. Check for an x⁻¹ term after simplifying.
06 / Expand before integrating
Integration does not distribute over multiplication.
First rewrite a product as a sum:
∫(x + 2)(2x − 1) dx
= ∫(2x² + 3x − 2) dx
= (2/3)x³ + (3/2)x² − 2x + C
Multiplying the two separate integrals is not a valid rule. Expanding makes the usual power rule available.
A square containing a root
(√x − 2)² = x − 4√x + 4
∫(√x − 2)² dx
= x²/2 − (8/3)x3/2 + 4x + C
The original square root requires x ≥ 0; differentiate the antiderivative on x > 0 and treat an endpoint separately.
07 / Compare unknown coefficients
An integral identity can be checked by differentiation.
Suppose, on x > 0:
∫[a/(2x³) − ab] dx = −3/(2x²) + 12x + C
Differentiate the right-hand side to get 3/x³ + 12. Matching the x⁻³ coefficients gives a/2 = 3, so a = 6. Matching the constant terms gives −ab = 12, so b = −2.
Alternatively integrate the left side and compare coefficients of the same powers. Keep arbitrary C separate from a fixed linear term such as 12x.
08 / Rearrange a relation first
Find an explicit real expression for y.
If √y = ∛x + 2 with x ≥ 0, square the complete right-hand side:
y = (∛x + 2)²
= x2/3 + 4x1/3 + 4
∫y dx = (3/5)x5/3 + 3x4/3 + 4x + C
The condition x ≥ 0 makes the right side of the original square-root equation non-negative, so squaring introduces no sign conflict here. For a general equation y² = g(x), choose the stated positive or negative branch before integrating.
09 / Integrate a finite approximation
State which polynomial terms were omitted.
The exact expansion of (1 − 2x)⁴ is 1 − 8x + 24x² − 32x³ + 16x⁴. If F′(x) = (1 − 2x)⁴ and F(0) = 0, then:
F(x) = x − 4x² + 8x³ − 8x⁴ + (16/5)x⁵
For small |x|: F(x) ≈ x − 4x² + 8x³
The omitted terms in F start at x⁴. The initial condition fixes the constant; without it, retain + C. This conclusion uses known polynomial terms, not an unspecified numerical approximation.
10 / Your turn
Differentiate each answer to check it.
Include C and any original domain restrictions.
02 · A polynomial
Find ∫(9x² − 6x + 2) dx.
Hint
The final constant term becomes linear.
Worked solution
3x³ − 3x² + 2x + C.
03 · A reciprocal
Find ∫4/x³ dx on an interval excluding zero.
Hint
Use 4x⁻³.
Worked solution
−2x⁻² + C = −2/x² + C.
04 · Fractional powers
Find ∫(2√x + 3/√x) dx, x > 0.
Hint
The powers are 1/2 and −1/2.
Worked solution
(4/3)x3/2 + 6√x + C.
05 · A product
Find ∫(x − 1)(x + 4) dx.
Hint
Expand to x² + 3x − 4.
Worked solution
x³/3 + 3x²/2 − 4x + C.
06 · Split the numerator
Find ∫(2x² + 5)/x² dx, x ≠ 0.
Hint
Rewrite as 2 + 5x⁻².
Worked solution
2x − 5/x + C, on either interval excluding zero.
07 · Another variable
Find ∫(rt³ − s) dt, treating r and s as constants.
Hint
Only t changes.
Worked solution
(r/4)t⁴ − st + C.
08 · Root brackets
Find ∫√x(√x + 1)² dx for x > 0.
Hint
The integrand expands to x3/2 + 2x + x1/2.
Worked solution
(2/5)x5/2 + x² + (2/3)x3/2 + C.
09 · A forbidden exponent
Why can’t the power rule above evaluate ∫1/x dx?
Hint
What is n + 1?
Worked solution
Here n = −1, making the denominator n + 1 zero. A different integration rule is needed; division by zero is not an answer.
10 · An identity
If ∫(ax² + b) dx = 2x³ − 5x + C, find a and b.
Hint
Differentiate the right-hand side.
Worked solution
ax² + b = 6x² − 5
a = 6, b = −5.
11 · A root relation
If √y = ∛x + 1 with x ≥ 0, find ∫y dx.
Hint
Square to get x2/3 + 2x1/3 + 1.
Worked solution
(3/5)x5/3 + (3/2)x4/3 + x + C.
12 · Binomial approximation
Integrate the first three terms of (1 + x)⁵ to approximate an antiderivative for small |x|.
Hint
The expansion starts 1 + 5x + 10x².
Worked solution
x + (5/2)x² + (10/3)x³ + C.
The omitted antiderivative terms start at x⁴. This is an approximation, not the exact integral of the full expression.
11 / Recap
Reverse the rule, then check the derivative.
- Increase the exponent and divide by the new exponent; n = −1 is excluded.
- Integrate constants as linear terms and sums term by term.
- Expand or simplify products, roots and fractions first.
- Read dx or dt carefully and treat other parameters as constants.
- Include C and retain the original domain.
- Differentiate your result to recover the integrand.
Section 1 of 11 · Reverse differentiation