Evaluate $\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x$. Latest Edit
By the contributions of the writers, we finally get the closed form for the integral as:
$$\boxed{\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x =-\frac{\pi(2 n-3) ! !}{2^{n}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}}$$

I first evaluate $$I_1=\int_{0}^{\infty} \frac{\ln x}{x^{2}+1} d x \stackrel{x\mapsto\frac{1}{x}}{=}  -I_1 \Rightarrow I_1= 0.$$
and then start to raise up the power of the denominator
$$I_n=\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x .$$
In order to use differentiation, I introduce a more general integral
$$I_n(a)=\int_{0}^{\infty} \frac{\ln x}{(x^{2}+a)^n} d x. $$
Now we can start with $I_1(a)$. Using $I_1=0$ yields
$$\displaystyle  1_1(a)=\int_{0}^{\infty} \frac{\ln x}{x^{2}+a} d x \stackrel{x\mapsto\frac{x}{a}}{=} \frac{\pi \ln a}{4 \sqrt a} \tag*{}$$
Now we are going to deal with $I_n$ by differentiating it by $(n-1)$ times
$$
\frac{d^{n-1}}{d a^{n-1}} \int_{0}^{\infty} \frac{\ln x}{x^{2}+a} d x=\frac{\pi}{4} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{a}\right)
$$
$$
\int_{0}^{\infty} \ln x\left[\frac{\partial^{n-1}}{\partial a^{n-1}}\left(\frac{1}{x^{2}+a}\right)\right] d x=\frac{\pi}{4} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt a}\right)
$$
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+a\right)^{n}} d x=\frac{(-1)^{n-1} \pi}{4(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)
$$
In particular, when $a=1$, we get a formula for $$
\boxed{\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x=\left.\frac{(-1)^{n-1} \pi}{4(n-1)!} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)\right|_{a=1}}
$$
For example, $$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{5}} d x=\frac{\pi}{4 \cdot 4 !}(-22)=-\frac{11 \pi}{48}
$$
and
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{10}} d x=\frac{-\pi}{4(9 !)}\left(\frac{71697105}{256}\right)=-\frac{1593269 \pi}{8257536}
$$
which is check by WA.
MY question
Though a formula for $I_n(a)$ was found, the last derivative is hard and tedious.
Is there any formula for $$\frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)? $$

Alternative solution:
Using the identity ($q > 0$)
$$
\frac{1}{q^n} = \frac{1}{\Gamma(n)}\int_0^\infty \mathrm{e}^{-qy}y^{n - 1}\mathrm{d} y,
$$
we have
\begin{align*}
 I_n &= \int_0^\infty \frac{\ln x}{(x^2 + 1)^n}\mathrm{d} x\\
 &= \int_0^\infty  \left( \frac{1}{\Gamma(n)}\int_0^\infty e^{-y(1 + x^2)}y^{n - 1}\mathrm{d}y \right) \ln x\, \mathrm{d} x \\
 &= \frac{1}{\Gamma(n)}\int_0^\infty \mathrm{e}^{-y} y^{n - 1}
 \left(\int_0^\infty \mathrm{e}^{-y x^2}\ln x \, \mathrm{d} x\right) \mathrm{d} y\\
 &= \frac{1}{\Gamma(n)}\int_0^\infty \mathrm{e}^{-y} y^{n - 1}
 \left( - \frac{(\gamma + \ln 4)\sqrt \pi}{4\sqrt y} - \frac{\sqrt{\pi}\ln y}{4\sqrt y} \right) \mathrm{d} y \\
 &= -\frac{(\gamma + \ln 4)\sqrt \pi}{4\Gamma(n)}\int_0^\infty \mathrm{e}^{-y} y^{n - 3/2}\mathrm{d} y - \frac{\sqrt{\pi}}{4\Gamma(n)}
 \int_0^\infty \mathrm{e}^{-y} y^{n - 3/2}\ln y \, \mathrm{d}y\\
 &= -\frac{(\gamma + \ln 4)\sqrt \pi}{4\Gamma(n)}\Gamma(n - 1/2) - \frac{\sqrt{\pi}}{4\Gamma(n)}\Gamma'(n - 1/2)\\
 &= -\frac{(\gamma + \ln 4)\sqrt \pi}{4\Gamma(n)}\Gamma(n - 1/2) - \frac{\sqrt{\pi}}{4\Gamma(n)} \psi(n - 1/2)\Gamma(n - 1/2)
\end{align*}
where $\psi(\cdot)$ is the digamma function defined by $\psi(u) = \frac{\mathrm{d} \ln \Gamma(u)}{\mathrm{d} u} = \frac{\Gamma'(u)}{\Gamma(u)}$,
and we have used
\begin{align*}
 &\int_0^\infty \mathrm{e}^{-y x^2}\ln x \, \mathrm{d} x\\
 =\,& \int_0^\infty \frac{1}{\sqrt y}\mathrm{e}^{-u^2}\ln \frac{u}{\sqrt y}\,\mathrm{d} u \\
 =\,& \frac{1}{\sqrt y}\int_0^\infty \mathrm{e}^{-u^2}\ln u \,\mathrm{d} u - \frac{\ln y}{2\sqrt y}\int_0^\infty \mathrm{e}^{-u^2}\mathrm{d} u\\
 =\,& - \frac{(\gamma + \ln 4)\sqrt \pi}{4\sqrt y} - \frac{\sqrt{\pi}\ln y}{4\sqrt y},
\end{align*}
and
$$\int_0^\infty \mathrm{e}^{-y} y^{n - 3/2}\ln y \, \mathrm{d}y = \Gamma'(n - 1/2) = \psi(n - 1/2)\Gamma(n - 1/2).$$
