Suppose $q \in [0,1]$ with continuous distribution $\Gamma$; $x,y$ are non-negative.
$$ \max_{I(\cdot)} \int_0^1 \log(1 + y + q - I(q))\, d\Gamma(q)\\ \text{subject to}\\ \int_0^1 \log(1 + x + I(q))\, d\Gamma(q) \geq w. $$
How can I solve the problem? Under regularity conditions I can form the Lagrangian and obtain the first order conditions, but I cannot solve the integrals to obtain a closed form for $I(q)$.
Let $\zeta$ be the Lagrange multiplier. Then the optimal solution $I(q)$ must satisfy $$ -\int_0^1 (1 + y + q - I(q))^{-1} \, d\Gamma(q) + \zeta \int_0^1 (1 + x + I(q))^{-1} \, d\Gamma(q) = 0\\ \Leftrightarrow \int_0^1 \left(\zeta (1 + x + I(q))^{-1} - (1 + y + q - I(q))^{-1}\right) \, d\Gamma(q) = 0. $$ A sufficient condition for the last equality to hold is $$ \zeta (1 + x + I(q))^{-1} = (1 + y + q - I(q))^{-1}\\ \Leftrightarrow I(q) = \frac{-(1+x) + (1 + y + q)\zeta}{1 + \zeta} $$ In addition, the optimal solution $I(q)$ must satisfy $$ \int_0^1 \log(1 + x + I(q)) \, d\Gamma(q) \geq w, $$ for which a sufficient condition is that $$ \log(1 + x + I(q)) = w. $$ Then $$ \zeta = \frac{e^w}{2-e^w+q+x+y}. $$ Therefore, $$ I(q) = e^w-x-1. $$