a) \( \int_{1}^{2} \frac{4 x^{5}-3 x^{4}+x^{3}-1}{x^{2}} d x \)

To evaluate the integral \( \int_{1}^{2} \frac{4 x^{5}-3 x^{4}+x^{3}-1}{x^{2}} d x \), we can first simplify the integrand by dividing each term in the numerator by \( x^{2} \): \[ \int_{1}^{2} \left( 4x^{3} - 3x^{2} + x - \frac{1}{x^{2}} \right) dx. \] Now, we can integrate each term separately: 1. \( \int 4x^{3} dx = x^{4} + C \) 2. \( \int -3x^{2} dx = -x^{3} + C \) 3. \( \int x dx = \frac{x^{2}}{2} + C \) 4. \( \int -\frac{1}{x^{2}} dx = \frac{1}{x} + C \) Putting it all together, we have: \[ \int \left( 4x^{3} - 3x^{2} + x - \frac{1}{x^{2}} \right) dx = x^{4} - x^{3} + \frac{x^{2}}{2} + \frac{1}{x} + C. \] Now we evaluate this from \( x = 1 \) to \( x = 2 \): \[ \left[ x^{4} - x^{3} + \frac{x^{2}}{2} + \frac{1}{x} \right]_{1}^{2}. \] Calculating at \( x = 2 \): \[ 2^{4} - 2^{3} + \frac{2^{2}}{2} + \frac{1}{2} = 16 - 8 + 2 + \frac{1}{2} = 10 + \frac{1}{2} = 10.5. \] Now at \( x = 1 \): \[ 1^{4} - 1^{3} + \frac{1^{2}}{2} + 1 = 1 - 1 + \frac{1}{2} + 1 = \frac{1}{2} + 1 = 1.5. \] Subtracting the two results: \[ 10.5 - 1.5 = 9. \] Thus, the value of the integral is \[ \int_{1}^{2} \frac{4 x^{5}-3 x^{4}+x^{3}-1}{x^{2}} d x = 9. \]