Q:

$s(t) = \int_0^{t} \bigg(\sqrt{q^2 + q^4 + \frac{4}{9}q^6}\bigg) dq$

What's the technique to integrate this type of integral:
$$s(t) = \int_{0}^{t}  \bigg(\sqrt{q^2 + q^4 + \frac{4}{9}q^6}\bigg) dq$$

A:

Factor $\frac{2q}3{}$ out of the radical. Then complete the square on the fourth degree polynomial under the radical.
Next, make a $\tan\theta$ substitution, turning the integral into the integral of $\sec^3\theta$, which is a standard integral.

A:

Let $x=q^2$,
$$s(t) = \int_{0}^{t} \sqrt{q^2 + q^4 + \frac{4}{9}q^6}\> dq
=\frac12\int_{0}^{x_t}\sqrt{1+x+\frac49x^2}\> dx $$
Then, let $u=\frac{8x+9}{3\sqrt{7}}$
$$s(t)=\frac{21}{64}\int_{\frac3{\sqrt7}}^{u_t}\sqrt{1+u^2}\> du$$
and $u=\sinh w$, 
$$s(t) =\frac{21}{64}\int_{\sinh^{-1}\frac3{\sqrt7}}^{w_t}
\cosh^2 wdw =\frac{21}{128}(u_t\sqrt{1+u_t^2}-\frac{12}7+\sinh^{-1}u_t-\sinh^{-1}\frac3{\sqrt7})$$
where $u_t=\frac{8t^2+9}{3\sqrt{7}}$.

