Answer
$$ 8\left(\frac{1}{3}\left(\frac{\sqrt{4+x^2}}{2}\right)^3 -\frac{\sqrt{4+x^2}}{2}\right)+c$$
Work Step by Step
Given
$$\int\frac{x^3dx}{\sqrt{ x^2+4}} $$
Let $x=2\tan \theta \ \Rightarrow \ dx=2\sec^2 \theta d \theta$, then
\begin{align*}
\int\frac{x^3dx}{\sqrt{ x^2+4}} &=\int\frac{16\tan^3\theta \sec^2 \theta d \theta}{\sqrt{ 4\tan^2 \theta+4}} \\
&=\int\frac{16\tan^3\theta \sec^2 \theta d \theta}{2\sec\theta} \\
&=8\int \tan^3\theta \sec \theta d \theta\\
&=8\int (\sec^2\theta-1)\tan \theta \sec \theta d \theta\\
&= 8\left(\frac{1}{3}\sec^3\theta-\sec \theta\right)+c\\
&= 8\left(\frac{1}{3}\left(\frac{\sqrt{4+x^2}}{2}\right)^3 -\frac{\sqrt{4+x^2}}{2}\right)+c
\end{align*}
