Answer
$\dfrac{1}{56}$
Work Step by Step
We solve the integral by using the Fundamental Theorem of Calculus.
$\int_{0}^1 \int_{2}^3 \dfrac{1}{(x+4y)^3} \ dx \ dy= -\dfrac{1}{2} \times \int_{0}^1 [(x+4y)^{-2}]_2^3 \ dy\\= -\dfrac{1}{2} \times [\dfrac{-(3+4y)^{-1}}{4}+\dfrac{(2+4y)^{-1}}{4}]_0^1 \\=-\dfrac{1}{2} \times [\dfrac{-(3+4(1))^{-1}}{4}+\dfrac{(2+4(1))^{-1}}{4}+\dfrac{3^{-1}}{4}-\dfrac{2^{-1}}{4}] \\=\dfrac{1}{56}$
