∫₀¹ (2x³ - x² + 2x + 2)/(x² + 1) dx = \na 3π/4\nb 3π/4\nc 19/8\nd 19π/24

Explanation:

Step1: Perform polynomial long - division

Divide $2x^{3}-x^{2}+2x + 2$ by $x^{2}+1$. We get $2x - 1+\frac{3}{x^{2}+1}$. So, $\int_{0}^{1}\frac{2x^{3}-x^{2}+2x + 2}{x^{2}+1}dx=\int_{0}^{1}(2x - 1+\frac{3}{x^{2}+1})dx$.

Step2: Integrate term - by - term

$\int_{0}^{1}(2x - 1+\frac{3}{x^{2}+1})dx=\int_{0}^{1}2xdx-\int_{0}^{1}1dx + 3\int_{0}^{1}\frac{1}{x^{2}+1}dx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$ and $\int\frac{1}{x^{2}+1}dx=\arctan(x)+C$, we have: $\int_{0}^{1}2xdx=x^{2}\big|{0}^{1}=1^{2}-0^{2}=1$, $\int{0}^{1}1dx=x\big|{0}^{1}=1 - 0 = 1$, and $3\int{0}^{1}\frac{1}{x^{2}+1}dx=3\arctan(x)\big|_{0}^{1}=3(\arctan(1)-\arctan(0))=3(\frac{\pi}{4}-0)=\frac{3\pi}{4}$.

Step3: Calculate the result

$1-1+\frac{3\pi}{4}=\frac{3\pi}{4}$.

Answer:

B. $\frac{3\pi}{4}$