Question:

Step 1: Differentiate the given equation with respect to $x$, treating $y$ as a constant. The given equation is $$ \int_1^{xy} f(t)\,dt = y\int_1^x f(t)\,dt + x\int_1^y f(t)\,dt $$ Differentiating both sides with respect to $x$: $$ y \cdot f(xy) = y \cdot f(x) + \int_1^y f(t)\,dt $$ Step 2: Substitute $x=1$ into the differentiated equation. Using $f(1)=3$: $$ y \cdot f(y) = y \cdot f(1) + \int_1^y f(t)\,dt $$ $$ y \cdot f(y) = 3y + \int_1^y f(t)\,dt $$ Step 3: Differentiate the resulting equation with respect to $y$. $$ \frac{d}{dy} (y \cdot f(y)) = \frac{d}{dy} \left( 3y + \int_1^y f(t)\,dt \right) $$ Using the product rule on the left side and the Fundamental Theorem of Calculus on the right side: $$ f(y) + yf'(y) = 3 + f(y) $$ Step 4: Solve the differential equation for $f(y)$. $$ yf'(y) = 3 $$ $$ f'(y) = \frac{3}{y} $$ Integrate both sides with respect to $y$: $$ f(y) = \int \frac{3}{y}\,dy $$ $$ f(y) = 3\ln|y| + C $$ Since the domain is $\mathbb{R}$, and $f(1)=3$, we consider $y>0$ for $\ln y$ to be defined in the context of the initial condition. Using the initial condition $f(1)=3$: $$ 3 = 3\ln(1) + C $$ $$ 3 = 0 + C $$ $$ C = 3 $$ Thus, $$ f(y) = 3\ln|y| + 3 $$ Replacing $y$ with $x$: $$ f(x) = 3\ln|x| + 3 $$