(a) To find the quotient and remainder when \(8x^3 + 4x^2 + 2x + 7\) is divided by \(4x^2 + 1\), perform polynomial division:
1. Divide the leading term \(8x^3\) by \(4x^2\) to get \(2x\).
2. Multiply \(2x\) by \(4x^2 + 1\) to get \(8x^3 + 2x\).
3. Subtract \(8x^3 + 2x\) from \(8x^3 + 4x^2 + 2x + 7\) to get \(4x^2 + 7\).
4. Divide \(4x^2\) by \(4x^2\) to get \(1\).
5. Multiply \(1\) by \(4x^2 + 1\) to get \(4x^2 + 1\).
6. Subtract \(4x^2 + 1\) from \(4x^2 + 7\) to get \(6\).
The quotient is \(2x + 1\) and the remainder is \(6\).
(b) The integral becomes:
\(\int_0^{\frac{1}{2}} \left(2x + 1 + \frac{6}{4x^2 + 1}\right) \, dx\)
Split the integral:
\(\int_0^{\frac{1}{2}} (2x + 1) \, dx + \int_0^{\frac{1}{2}} \frac{6}{4x^2 + 1} \, dx\)
1. \(\int_0^{\frac{1}{2}} (2x + 1) \, dx = \left[x^2 + x\right]_0^{\frac{1}{2}} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\)
2. \(\int_0^{\frac{1}{2}} \frac{6}{4x^2 + 1} \, dx = 3 \arctan(2x)\) evaluated from 0 to \(\frac{1}{2}\).
\(= 3 \left(\arctan(1) - \arctan(0)\right) = 3 \left(\frac{\pi}{4}\right) = \frac{3\pi}{4}\)
Combine results: \(\frac{3}{4} + \frac{3\pi}{4} = \frac{3}{4}(1 + \pi)\).
