Answer
$2 \pi( \ln 2-\dfrac{1}{2})$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$\int^1_{-1} \int^{\sqrt{1-y^2}}_{-\sqrt{1-y^2}} \ln(x^2+y^2+1) \space dx \space dy =\int_0^{2 \pi} \int_0^{1} r \ln (r^2+1) \space dr \space d\theta $
Need to set up $1+r^2 =a \implies r dr=\dfrac{da}{2}$
or, $=\int_0^{2 \pi} [\dfrac{\ln 2}{ 2}- (1/2) \int_1^{2} \dfrac{(a-1)da}{t}] d\theta $
or, $=4\int^{\pi/2}_0 \int^1_0 \ln(r^2+1) \space r \space dr \space d\theta $
or, $=2 \times \int^{\pi/2}_0(\ln 4-1)d\theta $
or, $=2 \pi( \ln 2-\dfrac{1}{2})$
