Why am I getting ∫(3x+1)^3dx as ((3x + 1)^12 )/18?

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Integration of ∫(3x+1)^3dx

I have no clue why I am getting the integration of $\int(3x+1)^3dx$ as $((3x + 1)^{12} )/18$. The differentiation of $(3x+1)^3$ is done correctly, but I'm still getting the wrong answer.

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Lets integrate

$$I=\int (3x+1)^3 dx.$$

Let $u=3x+1$ then $du=3dx$ (so $dx=\frac{1}{3} du$) thus

\begin{align} I&= \frac{1}{3} \int u^3 du\\ &= \frac{1}{12} u^4 + C \\ &=\frac{1}{12}(3x+1)^4+C \end{align}

here $C$ is the constant of integration.