How to find the formula for $\Gamma^{\prime}(m) \textrm{ and }\Gamma^{\prime \prime}(m)?$ In my post, I had found
$$\int_{0}^{\infty} \frac{\ln ^{2} x}{(1+x)^{2}} d x
=\frac{\pi^{2}}{3}.$$
Using similar technique, we can evaluate the general integral
$$I=\int_{0}^{\infty} \frac{\ln ^{2} x}{(1+x)^{m}} d x,$$
where $m \in N$ using the related integral
$$
\begin{aligned}
J(a) =&\int_{0}^{\infty} \frac{x^{a}}{(1+x)^{m}} d x
\\=& \int_{0}^{\infty} \frac{x^{(a+1)-1}}{(1+x)^{(a+1)+(m-a-1)} }d x \\
=& B(a+1, m-a-1) \\
=& \frac{\Gamma(a+1) \Gamma(m-a)}{\Gamma(m)} \\
=& \frac{1}{(m-1) !}\Gamma(a+1) \Gamma(m-a)
\end{aligned}
$$
$$
\begin{aligned}
\frac{\partial^{2}}{\partial a^{2}}J(a)=\frac{1}{(m-1) !}\left[\Gamma^{\prime \prime}(a+1)\Gamma(m-a)-2 \Gamma^{\prime}(a+1) \Gamma^{\prime}(m-a)+\Gamma^{\prime \prime}(m-a)\right]
\end{aligned}
$$
Back to our integral,
$$
\begin{aligned}
I&=\left.\frac{\partial^{2}}{\partial a^{2}} J(a)\right|_ {a=0} \\&=\frac{1}{(m-1) !}\left[\left(\Gamma^{\prime \prime}(1) \Gamma(m)-2 \Gamma ^{\prime} {(1)}\Gamma^{\prime} (m)\right.+\Gamma^{\prime \prime}(m)\right] \\
&=\frac{1}{(m-1) !}\left[\left(\gamma  ^{2}+\frac{\pi}{6}\right)(m-1)! +2 \gamma \Gamma^{\prime}(m)+\Gamma^{\prime \prime}(m)\right]
\end{aligned}
$$
My Question
How to find the formula for $\Gamma^{\prime}(m) \textrm{ and }\Gamma^{\prime \prime}(m)?$
 A: $$\Gamma'(m)=\Gamma (m)\, \psi ^{(0)}(m)\qquad \text{and}\qquad \Gamma''(m)=\Gamma (m) \,[\psi ^{(0)}(m)]^2+\Gamma (m)\, \psi ^{(1)}(m)$$
The problem is that, for $m=2$, your last formula leads to $\frac{\pi  (\pi+1 )}{6} $
In fact the antiderivative
$$I_m=\int \frac{\log^2(x)}{(1+x)^m}\,dx$$
$$\frac {I_m}x =2 \, _4F_3(1,1,1,m;2,2,2;-x)-2 \log (x) \,
   _3F_2(1,1,m;2,2;-x)+$$ $$\frac{\left((x+1)^m-(x+1)\right)  \log ^2(x)}{(m-1)\, x\,(x+1)^{m}}$$ where, as usual appear generalized hypergeometric functions.
Concerning the integrals
$$J_m=\int_0^\infty \frac{\log^2(x)}{(1+x)^m}\,dx=\frac{6 \left(H_{m-2}\right){}^2+6 \psi ^{(1)}(m-1)+\pi ^2}{6 (m-1)}$$ which also write
$$J_m=\frac{\pi ^2}{3 (m-1)}+\frac{6 \left(\left(H_{m-2}\right){}^2+\psi ^{(1)}(m-1)\right)-\pi ^2}{6 (m-1)}$$ where the second part of the expression is a rational number.
If $b_m=\frac{6 \left(\left(H_{m-2}\right){}^2+\psi ^{(1)}(m-1)\right)-\pi ^2}{6 (m-1)}$, they make the sequence
$$\left\{0,0,\frac{1}{3},\frac{1}{2},\frac{7}{12},\frac{5}{8},\frac{29}{45},\frac{469}{720}
   ,\frac{29531}{45360},\frac{1303}{2016},\frac{16103}{25200},\frac{190553}{302400},\frac
   {128977}{207900},\frac{9061}{14850},\cdots\right\}$$ This works for non-integar values of $m$.
We could do similar things changing the exponent of the logarithm; for example
$$K_m=\int \frac{\log^3(x)}{(1+x)^m}\,dx$$
$$\frac {K_m}x =-6 \, _5F_4(1,1,1,1,m;2,2,2,2;-x)-3 \log ^2(x) \, _3F_2(1,1,m;2,2;-x)+$$ $$6 \log (x) \,
   _4F_3(1,1,1,m;2,2,2;-x)+\frac{\left((x+1)^m-(x+1)\right) \log ^3(x)}{(m-1)\, x\,(x+1)^{m} }$$
