On the integral $I(a)=\int_0^1\frac{\log(a+t^2)}{1+t^2}\mathrm dt$
$$\mathcal J(a,t)=\int_0^1 \frac{\ln(a+t(1+x^2))}{1+x^2}\mathrm dx\Rightarrow I(a)=\int_0^1\frac{\ln(a+x^2)}{1+x^2}\mathrm dx=\mathcal J(a-1,1)$$ $$ \frac{\mathrm d}{\mathrm dt}\mathcal J(a,t)=\int_0^1 \frac{\mathrm dx}{a+t+tx^2}=\frac{1}{\sqrt{t(t+a)}}\arctan\left(\sqrt{\frac{t}{t+a}}\right)$$ $$\mathcal J(a,0)=\frac{\pi\ln a}{4}\Rightarrow \mathcal J(a,1)=\underbrace{\int_0^1 \frac{1}{\sqrt{t(t+a)}}\arctan\left(\sqrt{\frac{t}{t+a}}\right)\mathrm dt}_{=J}+\frac{\pi\ln a}{4}$$ Now via the substitution $\displaystyle{\sqrt{\frac{t}{t+a}}=x\Rightarrow \frac{\mathrm dt}{\sqrt{t(t+a)}}=\frac{2}{1-x^2}dx}$ we get: $$J=2\int_0^\frac{1}{\sqrt{1+a}}\frac{\arctan x}{1-x^2}\mathrm dx \overset{x=\frac{1-y}{1+y}}=\int_{\frac{\sqrt{1+a}-1}{\sqrt{1+a}+1}}^1\frac{\arctan\left(\frac{1-y}{1+y}\right)}{y}\mathrm dy$$ $$=\frac{\pi}{4}\int_{\frac{\sqrt{1+a}-1}{\sqrt{1+a}+1}}^1\frac{\mathrm dy}{y}-\int_0^1 \frac{\arctan y}{y}\mathrm dy+\int^{\frac{\sqrt{1+a}-1}{\sqrt{1+a}+1}}_0\frac{\arctan y}{y}\mathrm dy$$ $$\Rightarrow \mathcal J(a,1)=\frac{\pi}{4} \ln\left(\frac{\sqrt{a+1}+1}{\sqrt{a+1}-1}\right)-\mathrm G+\operatorname{Ti}_2\left(\frac{\sqrt{a+1}-1}{\sqrt{a+1}+1}\right)+\frac{\pi}{4}\ln a$$ $$\Rightarrow \boxed{I(a)=\int_0^1 \frac{\ln(a+x^2)}{1+x^2}dx=\frac{\pi}{2}\ln(\sqrt a+1)-\mathrm G+\operatorname{Ti}_2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)}$$ Where $\operatorname{Ti}_2(x)$ is the inverse tangent integral and $\mathrm G$ is Catalan's constant.
Some nice results that follows: $$\boxed{I(3)=\int_0^1\frac{\ln(3+x^2)}{1+x^2}\mathrm dx=\frac{\pi}{4}\ln 2+\frac{\pi}{6}\ln(2+\sqrt 3)-\frac13\mathrm G}$$ $$\boxed{I\left(\frac13\right)=\int_0^1 \frac{\ln\left(\frac13 +x^2\right)}{1+x^2}\mathrm dx=\frac{\pi}4 \ln \left(\frac23\right)+\frac{\pi}{3}\ln(2+\sqrt 3)-\frac53\mathrm G}$$
I get $$ I'(a) = \int_0^1 \dfrac{dt}{(t^2+a)(t^2+1)} = {\frac {\pi\,\sqrt {a}-4\,\arctan \left( {\frac {1}{\sqrt {a}}} \right) }{ 4 \left( a-1 \right) \sqrt {a}}} $$ Integrating this using Maple produces a rather complicated expression, which seems to work for $0 < a < 1$ (a different branch should be used after $a=1$): $$\frac{i}{2}{\it dilog} \left( {\frac {-2\,\sqrt {a}+1-i+ \left( 1+i \right) a }{a+1}} \right) -\frac{i}{2}{\it dilog} \left( {\frac {2\,\sqrt {a}+1+i+ \left( 1-i \right) a}{a+1}} \right) +\frac{\pi}{4}\,\ln \left( 1-\sqrt {a } \right) +\frac{\pi}{4}\,\ln \left( 1+\sqrt {a} \right) -\frac{1}{2}\,\arctan \left( {\frac {1}{\sqrt {a}}} \right) \ln \left( -\sqrt {2} \left( 1 +\sqrt {a} \right) \sqrt {a+1}+2\,a+2 \right) +\frac{1}{2}\,\arctan \left( { \frac {1}{\sqrt {a}}} \right) \ln \left( \sqrt {2} \left( -1+\sqrt {a } \right) \sqrt {a+1}+2\,a+2 \right) -\frac{1}{2}\,\arctan \left( {\frac {1}{ \sqrt {a}}} \right) \ln \left( \sqrt {2} \left( 1+\sqrt {a} \right) \sqrt {a+1}+2\,a+2 \right) +\frac{1}{2}\,\arctan \left( {\frac {1}{\sqrt {a}}} \right) \ln \left( -\sqrt {2} \left( -1+\sqrt {a} \right) \sqrt {a+1 }+2\,a+2 \right) +2\,i\arctan \left( {\frac {1}{\sqrt {a}}} \right) \arctan \left( {\frac {-1+\sqrt {a}}{-\sqrt {2}\sqrt {a+1}+\sqrt {a}+1 }} \right) +2\,i\arctan \left( {\frac {1}{\sqrt {a}}} \right) \arctan \left( {\frac {1+\sqrt {a}}{\sqrt {2}\sqrt {a+1}+\sqrt {a}-1}} \right) -i\arctan \left( {\frac {1}{\sqrt {a}}} \right) \pi-\frac{i}{4}{\pi}^ {2}-{\it Catalan} $$
Let $a>-1$ be a real number. Then
$$\int_0^1 \frac{\log(1+a^2x^2)}{1+x^2}\textrm{d}x=\frac{\pi}{2}\log(1+a)-G+\text{Ti}_2\left(\frac{1-a}{1+a}\right),$$
where $G$ is the Catalan's constant and $\displaystyle \text{Ti}_2(x)=\int_0^x \frac{\arctan(t)}{t}\textrm{d}t$ is the inverse tangent integral.
Thanks to Cornel for this way of writing the closed-form of the integral.