Split the integral at $v=1$ and let $v \to \frac{1}{v}$ in the second integral to obtain
$$ I_n \equiv - 2 \int \limits_0^\infty \frac{\sqrt{v} \ln^n (v)}{v^2 - 1} \, \mathrm{d} v = -2 \int \limits_0^1 \frac{[- \ln(v)]^n}{1-v^2} \left[v^{-1/2} + (-1)^{n-1} v^{1/2}\right] \, \mathrm{d} v \, . $$
Now let $v = \mathrm{e}^{-t}$ and expand the denominator into a geometric series:
$$ I_n = -2 \int \limits_0^\infty t^n \sum \limits_{l=0}^\infty \mathrm{e}^{-(2l+1)t} \left[\mathrm{e}^{t/2} + (-1)^{n-1} \mathrm{e}^{-t/2}\right] \, \mathrm{d} t \, . $$
Interchange summation and integration (monotone convergence theorem), make linear substitutions in the exponents and use the definition of the gamma function to find
$$ I_n = -2 n! \sum \limits_{l=0}^\infty \left[\frac{1}{\left(2l + \frac{1}{2}\right)^{n+1}} + \frac{(-1)^{n-1}}{\left(2l + \frac{3}{2}\right)^{n+1}}\right] = - 2^{n+2} n! \sum \limits_{m=0}^\infty \frac{(-1)^{(n-1) m}}{(2m+1)^{n+1}} \, .$$
For odd values of $n$ the sum is given by the Dirichlet lambda function:
\begin{align}
I_{2k-1} &= -2^{2k+1} (2k-1)! \lambda (2k) = -2^{2k+1} (2k-1)! \left(1-2^{-2k}\right) \zeta (2k) \\
&= - \frac{4^k (4^k-1) \lvert \mathrm{B}_{2k} \rvert}{2 k} \pi^{2k} \, , \, k \in \mathbb{N} \, .
\end{align}
$\mathrm{B}_{2k}$ are the Bernoulli numbers. For even $n$ one can use the Dirichlet beta function:
$$I_{2k} = - 2^{2k+2} (2k)! \beta(2k+1) = - \lvert \mathrm{E}_{2k} \rvert \pi^{2k+1} \, , \, k \in \mathbb{N} \, . $$
$\mathrm{E}_{2k}$ are the Euler numbers. In general, your integral is given by
$$ I_n = - a_n \pi^{n+1} \, , \, n \in \mathbb{N} \, ,$$
where $(a_n)_{n \in \mathbb{N}_0}$ is this sequence.
Note that these results suggest $I_0 = - \pi$, which is indeed true if we take the Cauchy principal value:
$$ I_0 \equiv - 2 \operatorname{PV} \int \limits_0^\infty \frac{\sqrt{v}}{v^2 - 1} \, \mathrm{d} v = -2 \int \limits_0^1 \frac{\mathrm{d}v}{\sqrt{v} (1+v)} = - 4 \int \limits_0^1 \frac{\mathrm{d}w}{1+w^2} = - \pi \, . $$
A large portion of this answer is taken directly from this answer!
Enforce the substitution
$$y=\sqrt{1-2x^2}\qquad\qquad x^2=\frac {1-y^2}{2}\qquad\qquad\mathrm dx=-\frac {y}{\sqrt{2(1-y^2)}}\,\mathrm dy$$
The integral now becomes
$$\begin{align*}\int\limits_0^{1/\sqrt3}\frac {\arctan\left(\frac 1{\sqrt{1-2x^2}}\right)}{1+x^2}\,\mathrm dx & =\sqrt{2}\int\limits_{1/\sqrt{3}}^1\frac {y\left(\frac {\pi}2-\arctan y\right)}{\sqrt{1-y^2}(3-y^2)}\,\mathrm dy\\ & =\frac {\pi}{\sqrt{2}}\int\limits_{1/\sqrt{3}}^1\frac {y\,\mathrm dy}{\sqrt{1-y^2}(3-y^2)}-\sqrt{2}\int\limits_{1/\sqrt3}^1\frac {y\arctan y}{\sqrt{1-y^2}(3-y^2)}\,\mathrm dy\end{align*}$$
Splitting up the integrand, the first integral can be evaluated with the substitution $y\mapsto\sqrt{1-y^2}$, giving
$$\frac {\pi}{\sqrt{2}}\int\limits_{1/\sqrt{3}}^1\frac {y}{\sqrt{1-y^2}(3-y^2)}\,\mathrm dy=\frac {\pi^2}{12}$$
And enforcing the substitution $t=\sqrt{y}$ on the second integral, we have that
$$\int\limits_0^{1/\sqrt3}\frac {\arctan\left(\frac 1{\sqrt{1-2x^2}}\right)}{1+x^2}\,\mathrm dx=\frac {\pi^2}{12}-\frac 1{\sqrt{2}}\int\limits_{1/3}^1\frac {\arctan\sqrt{t}}{(3-t)\sqrt{1-t}}\,\mathrm dt$$
The last integral is difficult, but using the same methodology and formulas as the answer I have linked above, we first rewrite the integral such that the lower limit is zero.
$$\int\limits_{1/3}^1\frac {\arctan\sqrt t}{(3-t)\sqrt{1-t}}\,\mathrm dt=\underbrace{\int\limits_0^1\frac {\arctan\sqrt{t}}{(3-t)\sqrt{1-t}}\,\mathrm dt}_{J}-\underbrace{\int\limits_0^{1/3}\frac {\arctan\sqrt{t}}{(3-t)\sqrt{1-t}}\,\mathrm dt}_{K}$$
Next, we use the following formula to evaluate the right-hand side
$$\int\limits_0^x\frac {\arctan\sqrt t}{(a-t)\sqrt{b-t}}\,\mathrm dt=\frac 1{\sqrt{a-b}}S\left(\arctan\sqrt{\frac {b-x}{a-b}},\arctan\sqrt{\frac {b+1}{a-b}},\arctan\frac 1{\sqrt{a}}\right)$$
There are two important observations that will help us evaluate the two integrals, namely
- $S(0,\beta,\gamma)=\pi(\beta-\gamma)$
- When $\sin^2\alpha+\sin^2\gamma=\sin^2\beta$, then $S(\alpha,\beta,\gamma)=-\alpha^2+\beta^2-\gamma^2$
Substituting $a=3$ and $b=1$, then
$$\begin{align*}J & =\frac 1{\sqrt2}S\left(0,\frac {\pi}4,\frac {\pi}6\right)=\frac {\pi^2}{12\sqrt2}\\K & =\frac 1{\sqrt2}S\left(\frac {\pi}6,\frac {\pi}4,\frac {\pi}6\right)=\frac {\pi^2}{144\sqrt2}\end{align*}$$
Where I have used the first observation to evaluate $J$ and the second observation to evaluate $K$. Taking the difference $J-K$, then
$$\int\limits_{1/3}^1\frac {\arctan x}{(3-x)\sqrt{1-x}}\,\mathrm dx=\frac {11\pi^2}{144\sqrt2}$$
Putting everything together, we get that
$$\int\limits_0^{1/\sqrt{3}}\frac {\arctan\left(\frac 1{\sqrt{1-2x^2}}\right)}{1+x^2}\,\mathrm dx=\frac {\pi^2}{12}-\frac {11\pi^2}{288}\color{blue}{=\frac {13\pi^2}{288}}$$
Best Answer
Note
\begin{align} \frac{I_1}{I_2}&=1+ \frac{I_1-I_2}{I_2} \\ &=1+\frac1{I_2}\int_0^{\frac\pi2} \left( \sqrt{1+\frac{1}{\sqrt{1+\tan^nx}}}- \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}} \right)dx \\ &=1+\frac1{I_2}\int_0^{\frac\pi2} \frac{\sqrt{\sqrt{\sin^nx+\cos^nx}+\sqrt{\cos^nx}} - \sqrt{\sqrt{\sin^nx+\cos^nx}-\sqrt{\cos^nx}}}{\sqrt[4]{\sin^nx+\cos^nx}}dx \\ &=1+\frac1{I_2} \int_0^{\frac\pi2} \frac{\sqrt{2\sqrt{\sin^nx+\cos^nx}-2\sqrt{\sin^nx}} }{\sqrt[4]{\sin^nx+\cos^nx}}dx \\ &= 1+\frac{\sqrt2}{I_2} \int_0^{\frac\pi2} \sqrt{1-\frac{1}{\sqrt{1+\cot^nx}}}dx\>\>\>\>\>\>\>\>(x\to \frac\pi2-x)\\ &=1+\frac{\sqrt2}{I_2}\int_0^{\frac\pi2} \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}}dx \\ &=1+\sqrt2 \\ \end{align}