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

Use $(a^x)’=a^x\ln a$ to evaluate
\begin{align}\frac{d^{n}}{d a^{n}}\left(\frac{\ln a}{\sqrt{a}}\right)_{a=1}
= &\frac{\partial^{n}}{\partial a^{n}}
\left(\frac{\partial a^x}{\partial x}\bigg|_{x=-\frac12}\right)_{a=1}
 =\frac{\partial}{\partial x}
\left(\frac{\partial^{n}a^x}{\partial a^{n}}\bigg|_{a=1}\right)_{x=-\frac12}\\
=& \frac{d}{dx}
\bigg(\prod_{k=1}^{n}(x-k+1)\bigg)_{x=-\frac12}
= \prod_{k=1}^{n}(\frac12-k)\sum_{j=1}^{n}\frac1{\frac12-j}\\
=&-\frac{(-1)^n (2n-1)!!}{2^{n-1}}
 \sum_{j=1}^{n}\frac1{2j-1}
\end{align}
Thus
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n+1}} d x=\frac{(-1)^{n} \pi}{4n!} \frac{d^{n}}{d a^{n}}\left(\frac{\ln a}{\sqrt{a}}\right)_{a=1}
 =-\frac{\pi(2n-1)!!}{2^{n+1}n!}
 \sum_{j=1}^{n}\frac1{2j-1}
$$
