Can a certain series of integrals over $[0,\frac{1}{16}]$ be solved using integration-by-parts? I have a series of six integrals. The first, say, is 
\begin{equation}
\int_ 0^{\frac {1} {16}} - 
 48 q^{3/2}\sqrt {18 q - 2\sqrt {17 q + 4}\sqrt {q} + 4} \mbox {d} q.
\end{equation}
Mathematica returns this as unperformed, as well as the subproblem
\begin{equation}
\int_ 0^{\frac {1} {16}} \sqrt {18 q - 2\sqrt {17 q + 4}\sqrt {q} + 4} \mbox {d} q.
\end{equation}
However, high-precision numerical integration shows that the latter integral equals $\frac{25}{204}$.
So, is this last result indicative that the original integral can be solved, in one way of another, by integration-by-parts? (I don't immediately see how.)
The other five integrands in my series--all exhibiting similar numerical phenomenon (upon division by $\frac{25}{204}$, after integration over $[0,\frac{1}{16}]$)--are 
\begin{equation}
\left\{-\frac{2352}{5} q^{5/2} \sqrt{18 q-2 \sqrt{17 q+4} \sqrt{q}+4},-\frac{96}{5}
   \sqrt{2} \sqrt{q} \sqrt{9 q-\sqrt{17 q+4} \sqrt{q}+2},-144 \sqrt{2} q^{3/2} \sqrt{9
   q-\sqrt{17 q+4} \sqrt{q}+2},\frac{96}{5} \sqrt{2} q \sqrt{(-17 q-4) \left(-9 q+\sqrt{q
   (17 q+4)}-2\right)},\frac{432}{5} \sqrt{2} q^2 \sqrt{(-17 q-4) \left(-9 q+\sqrt{q (17
   q+4)}-2\right)}\right\}.
\end{equation}
Needless to say, I am interested in the evaluation of the six integrals through any appropriate methodology. (Actually, I am specifically interested in the sum of the six integrals, more so  than each one itself.)
 A: Since $q>0$, let's rewrite the integral expression as:
$$
\sqrt{17q+4 - 2\sqrt{17q+4}\sqrt{q} + q} = \sqrt{\left(\sqrt{17q+4}-\sqrt{q}\right)^2}=\sqrt{17q+4}-\sqrt{q}.
$$
This expression is way easier to integrate. 
Another problem part in your integrals is actually the same:
$$
9q - \sqrt{17q+4}\sqrt{q} + 2=\frac12\left(\sqrt{17q+4}-\sqrt{q}\right)^2.
$$
I hope you will figure out everything else.
A: Yes, employing the transformations indicated by Vasily Mitch, the results of the six integrations are 
\begin{equation}
\left\{\frac{1133}{9248}-\frac{384 \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)}{289
   \sqrt{17}},\frac{147 \left(655360 \sqrt{17}
   \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)-1431757\right)}{855255040},\frac{3
   \left(256 \sqrt{17}
   \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)-901\right)}{2890},\frac{3399}{9248}-\frac{1152 \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)}{289
   \sqrt{17}},\frac{88043-12288 \sqrt{17}
   \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)}{393040},\frac{1728
   \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)}{4913
   \sqrt{17}}-\frac{593721}{50309120}\right\}.
\end{equation}
Summing these six results, together with another seven (earlier obtained without the need for the transformations), gives me my desired ("bound-entanglement") probability result of 
\begin{equation}
\frac{24 \text{csch}^{-1}\left(\frac{8}{\sqrt{17}}\right)}{17 \sqrt{17}}-\frac{91}{544} \approx 0.002187222381.
\end{equation}
So, to get back to the title of my question, "Can a certain series of integrals over $[0,\frac{1}{16}]$ be solved using integration-by-parts?"--perhaps the answer is in the negative, but moot, in any case. 
Serendipity!
