Denoting $z=\sin^2 a;\,a\in[0;\frac\pi2]$
$$I(z)=\int_0^1\frac{\log\vert z-t\vert}{\sqrt{t-t^2}}dt\overset{x=\sqrt t}{=}2\int_0^1\frac{\ln|z-x^2|}{\sqrt{1-x^2}}dx$$
$$\overset{x=\sin\phi}{=}2\int_0^\frac\pi2\ln|\sin^2a-\sin^2\phi|d\phi=\frac14\int_0^{4\pi}\ln|\sin^2a-\sin^2\phi|d\phi$$
$$=\frac14\int_0^{4\pi}\left(\ln\Big|2\sin\frac{\phi-a}2\,\cos\frac{\phi+a}2\Big|+\ln\Big|2\sin\frac{\phi+a}2\,\cos\frac{\phi-a}2\Big|\right)d\phi$$
$$=\frac12\int_0^{2\pi}\Big(2\ln 2+\ln|\sin(\phi-a)|+\ln|\sin(\phi+a)|+\ln|\cos(\phi-a)|+\ln|\cos(\phi+a)|\Big)d\phi$$
We see that $a$ (or, initially $z$) does not contribute due to periodicity of the functions of the integrand.
Taking into consideration that $\displaystyle \int_0^\frac\pi2\ln(\cos\phi)d\phi=\int_0^\frac\pi2\ln(\sin\phi)d\phi=-\,\frac\pi 2\ln2$
$$I(z)=2\pi\ln2+\frac12\cdot4\cdot4\int_0^\frac\pi2\ln(\cos\phi)d\phi=-2\pi\ln2$$