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

Your integral can be expressed as the derivative of the complete beta function:
From Fubini-Tonelli we can interchange limit/derviative and integral:
$$I = \int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x = \int_{0}^{\infty} \frac{\left(\frac{d}{dt}\Big|_{t=0+} x^t\right)}{(x^2+1)^n}dx = \frac{d}{dt}\Big|_{t=0+} \int_{0}^{\infty} \frac{x^t}{(1+x^2)^n} dx $$
Recall the following integral representation of the complete Beta function:
$$ \int_{0}^{\infty} \frac{y^{a-1}}{(1+y)^{a+b}} dx = B(a,b)$$
If we let $x^2 = w$ then
$$ I = \frac{d}{dt}\Big|_{t=0+} \frac{1}{2}\int_{0}^{\infty} \frac{w^{\frac{t-1}{2}}}{(1+w)^n}dw  = \frac{d}{dt}\Big|_{t=0+} \frac{1}{2} B\left(\frac{t+1}{2},n-\frac{t+1}{2}\right)$$
Hence
$$\lim_{t \to 0+} \frac{d}{dt} \frac{1}{2}
B\left(\frac{t+1}{2},n-\frac{t+1}{2}\right) = \frac{1}{4}\left[ \psi\left(\frac{1}{2}\right) - \psi\left(n-\frac{1}{2}\right)\right]B\left(\frac{1}{2},n-\frac{1}{2}\right) $$
where $\psi(z)$ is the digamma function.
The right hand side can also be expressed as
$$I= -\frac{1}{4}\left[ H_{n-\frac{3}{2}} + \ln(4)\right]B\left(\frac{1}{2},n-\frac{1}{2}\right) $$
where $H_{x}$ is the Harmonic number
Hence
$$\boxed{\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x = \frac{1}{4}\left[ \psi\left(\frac{1}{2}\right) - \psi\left(n-\frac{1}{2}\right)\right]B\left(\frac{1}{2},n-\frac{1}{2}\right) = -\frac{1}{4}\left[ H_{n-\frac{3}{2}} + \ln(4)\right]B\left(\frac{1}{2},n-\frac{1}{2}\right)}$$
Using this formula
$$ \psi\left(\frac{1}{2}-J\right) = \psi\left(\frac{1}{2}+J\right) = -\gamma -\ln(4) + \sum_{j=1}^{J} \frac{2}{2j-1} \quad J=0,1,2...$$
$$\psi\left(\frac{1}{2}\right) = -\gamma -\ln(4)$$
$$\psi\left(n-\frac{1}{2}\right) = -\gamma -\ln(4) + \sum_{j=1}^{n-1} \frac{2}{2j-1} $$
Hence
$$ \left[ \psi\left(\frac{1}{2}\right) - \psi\left(n-\frac{1}{2}\right)\right] = -\sum_{j=1}^{n-1} \frac{2}{2j-1}$$
We have another form:
$$ \boxed{\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x  = -\frac{1}{4}\sum_{j=1}^{n-1} \frac{2}{2j-1}B\left(\frac{1}{2},n-\frac{1}{2}\right) }$$
Similar to what other users found previously
