2. ∫(x² - x + 1)/x dx

Explanation:

Step1: Simplify the integrand

$\int\frac{x^{2}-x + 1}{x}dx=\int(\frac{x^{2}}{x}-\frac{x}{x}+\frac{1}{x})dx=\int(x - 1+\frac{1}{x})dx$

Step2: Integrate term - by - term

$\int(x - 1+\frac{1}{x})dx=\int xdx-\int 1dx+\int\frac{1}{x}dx$ We know that $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), $\int 1dx=x + C$ and $\int\frac{1}{x}dx=\ln|x|+C$. So, $\int xdx=\frac{x^{2}}{2}$, $\int 1dx=x$ and $\int\frac{1}{x}dx=\ln|x|$. Then $\int(x - 1+\frac{1}{x})dx=\frac{x^{2}}{2}-x+\ln|x|+C$

Answer:

$\frac{x^{2}}{2}-x+\ln|x|+C$