Help with integral $\int x^3\sqrt {1+4x^2}dx$ Id like to know how to solve the following integral
$\int x^3\sqrt {1+4x^2}dx$
I tried substituting $t=x^2$ but I dont know what to do from here.
 A: Hint: let $\color{red}{t=1+4x^2} \implies \frac{dt}{dx}=8x \iff \color{green}{xdx=\frac{1}{8}dt}$.
Then the integral becomes $$\begin{align}\int x^2 \sqrt{\color{red}{1+4x^2}} \color{green}{xdx} & =\color{green}{\frac{1}{8}}\int \color{purple}{x^2} \sqrt{\color{red}{t}} \color{green}{dt}\\ &={\frac{1}{8}}\int \color{purple}{\frac{t-1}{4}} \sqrt{t} dt \\
&=\frac{1}{32}\int(t-1)t^{1/2}dt\\&=\frac{1}{32}\int(t^{3/2}-t^{1/2})dt. \end{align}$$
You should be able to take it from here.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int x^{3}\root{1 + 4x^{2}}\,\dd x}$

\begin{align}&\color{#66f}{\large\int x^{3}\root{1 + 4x^{2}}\,\dd x}
=\half\
\overbrace{\int x^{2}\root{1 + 4x^{2}}\,\dd\pars{x^{2}}}^{\ds{\mbox{Set}\ x^{2} \equiv t\ \imp\ x = t^{1/2}}}\ =\
\half\int t\root{1 + 4t}\,\dd t
\\[3mm]&={1 \over 8}\int\bracks{\pars{1 + 4t}^{3/2} - \pars{1 + 4t}^{1/2}}\,\dd t
={1 \over 8}\bracks{%
{1 \over 10}\pars{1 + 4t}^{5/2} - {1 \over 6}\pars{1 + 4t}^{3/2}}
\\[3mm]&=\color{#66f}{\large%
{1 \over 240}\bracks{3\pars{1 + 4x^{2}}^{5/2} - 5\pars{1 + 4x^{2}}^{3/2}}}
+ \mbox{a constant}
\end{align}

