Evaluating two integrals involving $\tan^{-1}\left(\frac{\sqrt{x(1-x)}}{x+\frac12}\right)$

We can use an Euler substitution to simplify both integrals, namely: $\frac{\sqrt{x(1-x)}}{x}=t\Rightarrow x=\frac{1}{1+t^2}$. $$I=\int_0^1\arctan\left(\frac{\sqrt{x(1-x)}}{\frac{1}{2}+x}\right)dx=\int_0^\infty\arctan\left(\frac{2t}{3+t^2}\right)\left(\frac{1}{1+t^2}\right)'dt$$ $$\overset{IBP}=\int_0^\infty \left(\frac{1}{1+t^2}-\frac{3}{9+t^2}\right)\frac{1}{1+t^2}dt=\color{blue}{\int_0^\infty \frac{1}{(1+t^2)^2}dt}-\int_0^\infty \frac{1}{1+t^2}\frac{3}{9+t^2}dt$$ $$\overset{\color{blue}{t\to \frac{1}{t}}}=\color{blue}{\frac12\int_0^\infty \frac{1}{1+t^2}dt}+\frac{1}{8}\int_0^\infty \frac{3}{9+t^2}dt-\frac{3}{8}\int_0^\infty \frac{1}{1+t^2}dt=\frac{\pi}{8}$$

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Similarly for the second integral we obtain: $$J=\int_0^1\arctan\left(\frac{\sqrt{x(1-x)}}{\frac{1}{2}+x}\right)\frac{dx}{1-x}=2\int_0^\infty \arctan\left(\frac{2t}{3+t^2}\right) \frac{1}{t(1+t^2)}dt$$ $$\overset{IBP}=\int_0^\infty \left(\frac{1}{1+t^2}-\frac{3}{9+t^2}\right)(\ln(1+t^2)-2\ln t)dt$$ Now we can differentiate under the integral sign considering the following integral: $$J(a)=\int_0^\infty \left(\frac{1}{1+t^2}-\frac{3}{9+t^2}\right)(\ln(a^2+t^2)-2\ln t)dt$$ $$\Rightarrow J'(a)=\int_0^\infty \left(\frac{1}{1+t^2}-\frac{3}{9+t^2}\right)\frac{2a}{a^2+t^2}dt$$ $$=\int_0^\infty \left(\frac{1}{1-a^2}\frac{2a}{a^2+t^2}-\frac{2a}{1-a^2}\frac{1}{1+t^2}+\frac{2a}{9-a^2}\frac{3}{9+t^2}-\frac{3}{9-a^2}\frac{2a}{a^2+t^2}\right)dt$$ $$=\pi\left(\frac{1}{1-a^2}-\frac{a}{1-a^2}+\frac{a}{9-a^2}-\frac{3}{9-a^2}\right)=\pi\left(\frac{1}{1+a}-\frac{1}{3+a}\right)$$ $$J(0)=0\Rightarrow J=\pi\int_0^1 \left(\frac{1}{1+a}-\frac{1}{3+a}\right)da=\pi\ln\left(\frac{3}{2}\right)$$