Answer
$$\displaystyle{\int\frac{x^3}{\sqrt{x^2+4}}dx=\frac{\left(x^2+4\right)^\frac{3}{2}}{3}-4\left(x^2+4\right)^\frac{1}{2}}+C$$
Work Step by Step
$\displaystyle{I=\int\frac{x^3}{\sqrt{x^2+4}}dx}$
$\displaystyle \left[\begin{array}{ll} x=2\tan\theta & x^3=8\tan^3\theta \\ & \\ \frac{dx}{d\theta}=2\sec^2\theta & dx=2\sec^2\theta\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int\frac{8\tan^3\theta}{\sqrt{4\tan^2\theta+4}}2\sec^2\theta\ d\theta}\\
\displaystyle{I=\int\frac{8\tan^3\theta}{\sqrt{4\left(\tan^2\theta+1\right)}}2\sec^2\theta\ d\theta}\\
\displaystyle{I=\int\frac{8\tan^3\theta}{2\sec\theta}}\\
\displaystyle{I=\int\frac{8\times2\tan^3\theta\sec^2\theta}{2\sec\theta}d\theta}\\
\displaystyle{I=8\int\tan^3\theta\sec\theta\ d\theta}\\
\displaystyle{I=8\int\frac{\sin^3\theta}{\cos^4\theta}d\theta}$
$\displaystyle \left[\begin{array}{ll} u=\cos\theta & 1-u^2=\sin^2\theta \\ & \\ \frac{du}{d\theta}=-\sin\theta & d\theta=-\frac{du}{\sin\theta} \end{array}\right]$ Integration by parts
$\displaystyle{I=8\int\frac{\sin^2\theta\sin\theta}{\cos^4\theta}\times -\frac{du}{\sin\theta}}\\
\displaystyle{I=-8\int\frac{1-u^2}{u^4}\ du}\\
\displaystyle{I=-8\int u^{-4}-u^{-2}\ du}\\
\displaystyle{I=-8\left(-\frac{1}{3}u^{-3}+u^{-1}\right)}+C\\
\displaystyle{I=\frac{8}{3u^{3}}-\frac{8}{u}}+C\\
\displaystyle{I=\frac{8}{3{\cos}^{3}\theta}-\frac{8}{\cos\theta}}+C\\
\displaystyle{\cos\theta=\frac{2}{\sqrt{x^2+4}}}\\
\displaystyle{I=\frac{8}{3\left({\frac{2}{\sqrt{x^2+4}}}\right)^{3}}-\frac{8}{{\frac{2}{\sqrt{x^2+4}}}}}+C\\
\displaystyle{I=\frac{\left(x^2+4\right)^\frac{3}{2}}{3}-4\left(x^2+4\right)^\frac{1}{2}}+C$
