Answer
$2 \ln (2)-1$
Work Step by Step
Re-arrange the given integral as follows:
$\int_{0}^{1}[\ln |1+xy|]_{0}^{1}]dy=\int_{0}^{1}[\ln |1+y|-\ln|1+0|] dy$
This implies that
$\int_0^1 \ln |1+y| dy=[y\ln |1+y|-y+\ln|1+y|]_{0}^{1}$
or, $\ln 2-1+\ln 2-0=2 \ln (2)-1$
