Integrating logarithms

$\begin{aligned}\text{Let, } I&=\int_{0}^{\infty} \dfrac{\ln(1+x^2)}{1+x^2} dx\\\\ \text{Now, let }x&=\tan\theta,dx=\sec^{2}\theta \cdot d\theta\\ I&=\int_{0}^{\tfrac{\pi}{2}} \dfrac{\ln(1+\tan^{2}\theta)}{1+\tan^{2}\theta}\hspace{3mm}\sec^{2}\theta\cdot d\theta\\ &=\int_{0}^{\tfrac{\pi}{2}} \ln(1+\tan^{2}\theta)\hspace{2mm}d\theta\\ &=-2\int_{0}^{\tfrac{\pi}{2}} \ln(\cos\theta)\hspace{2mm}d\theta\\ &=-2\int_{0}^{\tfrac{\pi}{2}} \ln(\sin\theta)\hspace{2mm}d\theta\hspace{7mm}\color{#3D99F6}\small\int_{a}^{b}f(x)\hspace{2mm}dx=\int_{a}^{b}f(a+b-x)\hspace{2mm}dx\\\\ \implies 2I&=-2\int_{0}^{\tfrac{\pi}{2}} \ln(\cos\theta)\hspace{2mm}d\theta+-2\int_{0}^{\tfrac{\pi}{2}} \ln(\sin\theta)\hspace{2mm}d\theta\\ &=-2\int_{0}^{\tfrac{\pi}{2}}\ln\left(\dfrac{\sin2\theta}{2}\right)\hspace{2mm}d\theta\\ &=-2\int_{0}^{\tfrac{\pi}{2}}\ln(\sin2\theta)\hspace{2mm}d\theta+2\int_{0}^{\tfrac{\pi}{2}}\ln2\hspace{2mm}d\theta\\ &=-2\int_{0}^{\tfrac{\pi}{2}}\ln(\sin2\theta)\hspace{2mm}d\theta+\pi\ln2\\\\ \text{Let, t}&=2\theta,dt=2\cdot d\theta\\ \implies 2I&=-\int_{0}^{\pi}\ln(\sin t)\hspace{2mm}dt+\pi\ln2\\ &=-2\int_{0}^{\tfrac{\pi}{2}}\ln(\sin t)\hspace{2mm}dt+\pi\ln2\hspace{7mm}\color{#3D99F6}\small\sin(\pi-t)=\sin t,\text{ thus }\int_{0}^{\pi}\ln(\sin t)\hspace{2mm}dt=2\int_{0}^{\tfrac{\pi}{2}}\ln(\sin t)\hspace{2mm}dt\\\\ \implies 2I&=I+\pi\ln2\\ I&=\pi\ln2\\ 10A+11B&=10\cdot (1)+11\cdot (2)=\color{#EC7300}\boxed{\color{#333333}32}\end{aligned}$

Denote our integral by $\displaystyle F(n)=\int_0^\infty \dfrac{\ln(1+x^n)}{1+x^n}\; dx$ for $n\;:-\;\{n\in\mathbb{R}|n>1\}$ .

$\displaystyle \begin{aligned} F(n)&= \int_0^\infty \dfrac{\ln(1+x^n)}{1+x^n}\; dx \\ &= \dfrac{1}{n}\int_1^\infty \dfrac{\ln x}{x} (x-1)^{1/n-1}\; dx \quad\color{#3D99F6}{\text{Set } (1+x^n)\to\; x} \\ &= \dfrac{-1}{n} \int_0^1 t^{\dfrac{-1}{n}} (1-t)^{\dfrac{1}{n}-1}\ln t\; dt\quad\color{#3D99F6}{\text{Set }x\to\dfrac{1}{x}} \\ &= \dfrac{-1}{n}\dfrac{\partial}{\partial a}\left(\int_0^1 t^a (1-t)^b\;dt\right)_{a=\frac{-1}{n} , b=\frac{1}{n}-1} \\ &= \dfrac{-1}{n}\dfrac{\partial}{\partial a}\left[\beta(a,b)\right]_{a=1-\frac{1}{n},b=\frac{1}{n}} \\ &= -\dfrac{1}{n}\beta\left(1-\dfrac{1}{n},\dfrac{1}{n}\right)\left[\psi\left(1-\dfrac{1}{n}\right)+\gamma\right] \\ &= \dfrac{-1}{n}\times \dfrac{\Gamma\left(1-\dfrac{1}{n}\right)\Gamma\left(\dfrac{1}{n}\right)}\times \left(\psi\left(1-\dfrac{1}{n}\right)+\gamma\right) \\ &= \dfrac{-\pi}{\sin\left(\dfrac{\pi}{n}\right)n}\left(\psi\left(1-\dfrac{1}{n}\right)+\gamma\right) \quad\text{Since } \Gamma(x)\Gamma(1-x)=\dfrac{\pi}{\sin(\pi x)}\end{aligned}$

Put $n=2$ to get the answer using $\psi\left(\dfrac{1}{2}\right)=-2\ln 2-\gamma$ as $\boxed{\pi\ln 2}$ and the answer is $32$