<s>[INST] Evaluate $\int_0^{\sqrt{5}} \sqrt{4x^4 + 16 x^2} dx$ $$\int_0^{\sqrt{5}} \sqrt{4x^4 + 16 x^2} dx$$
The square root really got me confused here. I've tried using the standard trick with $x = x + \sqrt{x^2+4}$ but failed. That might still be the best approach.
What's the best way to solve this and how to do it?
 [/INST] Hint
\begin{align}
\int^\sqrt{5}_0\sqrt{4x^4+16x^2} \ dx
&=2\int^{\sqrt{5}}_0 x\sqrt{x^2+4} \ dx \\
\end{align}
Now let $x=2\tan{u}$, then $dx=2\sec^2{u} \ du$.
\begin{align}
2\int^{\sqrt{5}}_0 x\sqrt{x^2+4} \ dx 
&=8\int^{\arctan\left(\frac{\sqrt{5}}{2}\right)}_0 \tan{u}\sec^3{u} \ du\\
\end{align}
Use the substitution $y=\sec{u} \implies dy=\sec{u}\tan{u} \ du$. You should be able to take it from here.
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