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 $$\frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)? $$

Inspired by metamorphy, I do want to try his wonderful trick
$$
I_{n}=\frac{1}{4} F^{\prime} _n\left(\frac{1}{2}\right) \text { where } F_{n}(\alpha)=2 \int_{0}^{\infty} \frac{x^{2 \alpha-1}}{\left(x^{2}+1\right)^{2}} d x
$$
In my post, I had found $$ \displaystyle \int_{0}^{\infty} \frac{x^{r} d x}{\left(x^{m}+a\right)^{n}}=\frac{\pi}{m(n-1) !} \csc \frac{(r+1) \pi}{m} \prod_{j=1}^{n-1}\left(j-\frac{r+1}{m}\right) a^{-\left(n-\frac{r+1}{m}\right)}\tag*{} $$
Replacing $m=2, r=2 \alpha-1 $ yields
$$
\begin{aligned}
F_{n}(\alpha) &=\frac{2 \pi}{2(n-1) !} \csc \left(\frac{2 \alpha \pi}{2}\right) \prod_{j=1}^{n-1}\left(j-\frac{2 \alpha}{2}\right) \\
&=\frac{\pi}{(n-1) !} \csc (\alpha \pi) \prod_{j=1}^{n-1}(j-\alpha)
\end{aligned}
$$
By logarithmic differentiontin, we get
$$
\frac{F_{n}^{\prime}(\alpha)}{F_{n}(\alpha)}=-\pi \cot (\alpha \pi)-\sum_{j=1}^{n-1}\left(\frac{1}{j-\alpha}\right)
$$
Putting $\alpha=\frac{1}{2}$ yields
$$
\begin{aligned}
F_{n}^{\prime}\left(\frac{1}{2}\right) &=-F_{n}\left(\frac{1}{2}\right) \sum_{j=1}^{n-1} \frac{1}{j-\frac{1}{2}} \\
&=-\frac{\pi}{(n-1) !} \prod_{j=1}^{n-1}\left(j-\frac{1}{2}\right) \sum_{j=1}^{n-1} \frac{1}{j-\frac{1}{2}}\\& =-\frac{\pi(2 n-3) ! !}{2^{n-2}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}
\end{aligned}
$$
Now we can conclude that
$$\boxed{I_n=-\frac{\pi(2 n-3) ! !}{2^{n}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}}$$
