$$I=\int\limits_0^1\dfrac 1{1-y}.\dfrac 1{-2}\int\limits^{x=\sqrt y}_{x=y}\dfrac {-2x}{\sqrt{y-x^2}}dxdy$$

(Adjusted-2 in Numerator and denominator)

$Let \space I_1=\int\limits^{x=\sqrt y}_{x=y}\dfrac {-2x}{\sqrt{y-x^2}}dx$

$Let \space y-x^2=t\\ \therefore -2x\space dx=dt\\ when, x=\sqrt y\\ t=0\\ when \space x=y,t=y-y^2\\ \therefore I_1=\int\limits_{y-y^2}^0\dfrac 1{\sqrt t}dt$

$ =\begin {bmatrix} \sqrt t\\ \dfrac 12\\ \end{bmatrix}_{y-y^2}^0 $

$=[2\sqrt0-2\sqrt{y-y^2}]\\ \therefore I=\int\limits_0^1\dfrac 1{1-y}\Bigg(\dfrac {-1}2\Bigg)\times (-2\sqrt{y-y^2})dy\\ =\int\limits^{1}_0\dfrac {\sqrt{y(1-y)}}{(1-y)}dy\\ =\int\limits_0^1y^{1/2}\times (1-y)^{-\frac 12}dy\\ \text{By Beta Gamma Formula }\\ I=\beta\Bigg(\dfrac 32,\dfrac 12\Bigg)\\ =\dfrac {\Big)\overline{\dfrac 32}.\Big)\overline{\dfrac 12}}{)\overline2}\\ =\dfrac 12.\Big)\overline{\dfrac 12}\times \sqrt{\pi}=\dfrac \pi2$