1. **Problem Statement:** Find the increase in the voting population $N(t)$ during the next 3 years given the rate of change $$\frac{dN}{dt} = \frac{100t}{(1+t)^2}$$ and initial population $N(0)$. 2. **Formula and Explanation:** The increase in population from $t=0$ to $t=3$ is given by the definite integral of the rate of change: $$\Delta N = \int_0^3 \frac{100t}{(1+t)^2} dt$$ 3. **Integration Steps:** Rewrite the integral: $$\int_0^3 \frac{100t}{(1+t)^2} dt = 100 \int_0^3 \frac{t}{(1+t)^2} dt$$ Let $u = 1 + t$, then $du = dt$ and when $t=0$, $u=1$; when $t=3$, $u=4$. Rewrite $t = u - 1$: $$100 \int_1^4 \frac{u - 1}{u^2} du = 100 \int_1^4 \left(\frac{u}{u^2} - \frac{1}{u^2}\right) du = 100 \int_1^4 \left(\frac{1}{u} - u^{-2}\right) du$$ 4. **Integrate term-by-term:** $$100 \left[ \int_1^4 \frac{1}{u} du - \int_1^4 u^{-2} du \right] = 100 \left[ \ln|u| \Big|_1^4 - \left(-\frac{1}{u}\right) \Big|_1^4 \right]$$ 5. **Evaluate definite integrals:** $$= 100 \left[ \ln 4 - \ln 1 + \left(\frac{1}{1} - \frac{1}{4}\right) \right] = 100 \left[ \ln 4 + \frac{3}{4} \right]$$ 6. **Final answer:** The increase in population over 3 years is: $$\boxed{100 \left( \ln 4 + \frac{3}{4} \right) \text{ thousands}}$$ This means the population increases by approximately $100 \times (1.3863 + 0.75) = 213.63$ thousands in 3 years.