The problem asks to evaluate the definite integral $\int_{1}^{2} (x^2 - 3x) \, dx + \int_{1}^{2} (x^2 + 4x + 1) \, dx$.

1. Problem Description

The problem asks to evaluate the definite integral $\int_{1}^{2} (x^2 - 3x) \, dx + \int_{1}^{2} (x^2 + 4x + 1) \, dx$.

2. Solution Steps

We can combine the integrals since they have the same limits of integration.

$\int_{1}^{2} (x^2 - 3x) \, dx + \int_{1}^{2} (x^2 + 4x + 1) \, dx = \int_{1}^{2} [(x^2 - 3x) + (x^2 + 4x + 1)] \, dx$

Simplify the expression inside the integral:

$\int_{1}^{2} (2x^2 + x + 1) \, dx$

Now, we find the antiderivative of $2x^2 + x + 1$:

$\int (2x^2 + x + 1) \, dx = \frac{2}{3}x^3 + \frac{1}{2}x^2 + x + C$

Now, we evaluate the definite integral:

$\int_{1}^{2} (2x^2 + x + 1) \, dx = \left[ \frac{2}{3}x^3 + \frac{1}{2}x^2 + x \right]_{1}^{2}$

$= \left( \frac{2}{3}(2)^3 + \frac{1}{2}(2)^2 + (2) \right) - \left( \frac{2}{3}(1)^3 + \frac{1}{2}(1)^2 + (1) \right)$

$= \left( \frac{2}{3}(8) + \frac{1}{2}(4) + 2 \right) - \left( \frac{2}{3} + \frac{1}{2} + 1 \right)$

$= \left( \frac{16}{3} + 2 + 2 \right) - \left( \frac{2}{3} + \frac{1}{2} + 1 \right)$

$= \frac{16}{3} + 4 - \frac{2}{3} - \frac{1}{2} - 1$

$= \frac{14}{3} + 3 - \frac{1}{2}$

$= \frac{14}{3} + \frac{6}{2} - \frac{1}{2}$

$= \frac{14}{3} + \frac{5}{2}$

$= \frac{28}{6} + \frac{15}{6}$

$= \frac{43}{6}$

3. Final Answer

$\frac{43}{6}$