Integration: Question 1

Worked solution

Step 1: Integrate each term with the power rule

Split the integral into its two terms:

$\int f(x)\,dx = \int 6x^2\,dx + \int (4x+1)^3\,dx$

First term. The ordinary power rule $\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1}+c$:

$\int 6x^2\,dx = 6 \times \frac{x^3}{3} = 2x^3$

Second term. The $(ax+b)^n$ rule $\displaystyle\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)}+c$, with $a=4$, $n=3$:

$\int (4x+1)^3\,dx = \frac{(4x+1)^4}{4 \times 4} = \frac{(4x+1)^4}{16}$

Adding the two results (and including a single constant of integration):

$\int f(x)\,dx = 2x^3 + \frac{(4x+1)^4}{16} + c$

Step 2: Recompute independently by differentiating the candidate answer

A safe check is to differentiate the proposed antiderivative and confirm it returns $f(x)$.

$\frac{d}{dx}\left(2x^3\right) = 6x^2$

$\frac{d}{dx}\left(\frac{(4x+1)^4}{16}\right) = \frac{4(4x+1)^3 \times 4}{16} = \frac{16(4x+1)^3}{16} = (4x+1)^3$

Adding these: $6x^2 + (4x+1)^3 = f(x)$ exactly, so the antiderivative in Step 1 is confirmed correct.

Why the other options are wrong

• B: the first term keeps the coefficient $6$ unchanged instead of dividing it by the new power $3$; it should reduce to $2x^3$, not $6x^3$.
• C: the second term is divided only by the new power $n+1=4$, but not also by the coefficient $a=4$ inside the bracket. The correct denominator is $a(n+1)=16$, not $4$.
• D: the second term’s power was never raised from $3$ to $4$ before dividing.

Final answer

$\boxed{\int f(x)\,dx = 2x^3 + \frac{(4x+1)^4}{16} + c} \quad \text{(Option A)}$