Integrating a Bracket Raised to a Power

To integrate a linear bracket to a power, use the reverse chain rule: raise the power by one, divide by the new power, and divide by the derivative of the bracket. For $(ax+b)^n$ that extra division is by $a$.

(a) Find $\displaystyle\int (2x+1)^3\,dx$. (b) Hence evaluate $\displaystyle\int_0^1 (2x+1)^3\,dx$. [5]

The working

Part (a), reverse chain rule. Raise the power to $4$, divide by $4$, then divide by the derivative of the bracket ($\frac{d}{dx}(2x+1) = 2$): $\int (2x+1)^3\,dx = \frac{(2x+1)^4}{4 \times 2} + c = \frac{(2x+1)^4}{8} + c \quad \text{(M1, A1)}$

Part (b), evaluate between $0$ and $1$ (the constant cancels in a definite integral): $\left[\frac{(2x+1)^4}{8}\right]_0^1 = \frac{(2(1)+1)^4}{8} - \frac{(2(0)+1)^4}{8} \quad \text{(M1)}$ $= \frac{3^4}{8} - \frac{1^4}{8} = \frac{81}{8} - \frac{1}{8} = \frac{80}{8} = 10 \quad \text{(A1)}$

Where the marks are won and lost

• Divide by the bracket’s derivative ($2$) as well as by the new power ($4$), giving $\div 8$ overall. Missing the $\div 2$ is the classic slip.
• For the definite integral, substitute the limits into the bracket carefully: $(2(1)+1)^4 = 3^4 = 81$.
• No $+c$ is needed in a definite integral, it cancels, but keep it in the indefinite answer.

Common mistakes

• Forgetting to divide by $2$, giving $\frac{(2x+1)^4}{4}$.
• Expanding $(2x+1)^3$ first (slower and error-prone) instead of using the reverse chain rule.
• Arithmetic on the powers: $3^4 = 81$, not $12$.

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