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**Prove It** First note that $\displaystyle \displaystyle \begin{align*} \int{x^3\sqrt{x^2 + 38}\,dx} = \frac{1}{2}\int{x^2\sqrt{x^2 + 38}\,2x\,dx} \end{align*} $

Let $\displaystyle \displaystyle \begin{align*} u = x^2 + 38 \implies du = 2x\,dx \end{align*} $ and the integral becomes

$\displaystyle \displaystyle \begin{align*} \frac{1}{2}\int{x^2\sqrt{x^2 + 38}\,2x\,dx} &= \frac{1}{2}\int{\left(u - 38\right)\sqrt{u}\,du} \\ &= \frac{1}{2}\int{u^{\frac{3}{2}} - 38u^{\frac{1}{2}}\,du} \\ &= \frac{1}{2}\left(\frac{u^{\frac{5}{2}}}{\frac{5}{2 }} - \frac{38u^{\frac{3}{2}}}{\frac{3}{2}}\right) + C \\ &= \frac{1}{2}\left[\frac{2}{5}\left(x^2 + 38\right)^{\frac{5}{2}} - \frac{76}{3}\left(x^2 + 38\right)^{\frac{3}{2}}\right] + C \\ &= \frac{1}{5}\left(x^2 + 38\right)^{\frac{5}{2}} - \frac{38}{3}\left(x^2 + 38\right)^{\frac{3}{2}} + C \end{align*} $

So you were almost there, you just forgot to halve everything.