$-\Delta u = f(x)\in L^1(\Omega)\quad$ in $\Omega$
$u = 0\quad$ on $\partial \Omega$
since the term $\int_\Omega f(x)v(x)$ does not make sense, if $f\in L^1(\Omega), v\in W^{1,2}_0(\Omega)$;
-div$(\frac{\nabla u}{(1+|u|)^\theta})=f(x)\in L^2(\Omega)\quad$ in $\Omega$
$u = 0\quad$ on $\partial \Omega$
Since the term $\int_\Omega \frac{|\nabla v|^2}{(1+|v|)^\theta}$ goes to zero, if $v$ is large.
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