Answer
$\dfrac{\ln (2)}{3}$
Work Step by Step
Need to integrate the given integral first with respect to $x$, then $z$, and then $y$
$ \int_{0}^1 \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} \dfrac{z}{y+1} dx dz dy= \int_{0}^1 \int_{0}^{1} [\dfrac{xz}{y+1}]_{0}^{\sqrt{1-z^2}} dz dy$
or, $=\int_{0}^1 \dfrac{1}{y+1} dy \int_{0}^{1} z\sqrt{1-z^2} dz$
Plug $k=1-z^2 \implies z dz=-\dfrac{dk}{2}$
or, $=[\ln |y+1|]_{0}^1 \int_{0}^{1} \sqrt k [-\dfrac{dk}{2}]$
or, $= (\ln 2)(\dfrac{-1}{2}) \int_0^1 [k^{1/2}] dk$
or, $=(ln 2) (\dfrac{1}{3})$
or, $=\dfrac{\ln (2)}{3}$
