Answer
$\dfrac{\ln17}{4}$
Work Step by Step
$\int_{0}^{4}\int_{0}^{\sqrt{x}}\dfrac{y}{x^2+1}\,dy\,dx=\dfrac{1}{2}\int_{0}^{4}\biggl[\dfrac{y^2}{x^2+1}\biggr]_{0}^{\sqrt{x}}\,dx=\frac{1}{2}\int_{0}^{4}\dfrac{x}{x^2+1}\, dx=\frac{1}{4}\biggl[\ln(x^2+1)\biggr]_{0}^{4}=\dfrac{\ln17}{4}$
