To find the indefinite integral [tex]\(\int \frac{x^2 + x + 1}{\sqrt{x}} \, dx\)[/tex], we will start by simplifying the integrand. We can rewrite the integrand as follows:
[tex]\[
\frac{x^2 + x + 1}{\sqrt{x}} = \frac{x^2}{\sqrt{x}} + \frac{x}{\sqrt{x}} + \frac{1}{\sqrt{x}}
\][/tex]
Next, we simplify each term separately:
1. [tex]\(\frac{x^2}{\sqrt{x}} = x^{2 - \frac{1}{2}} = x^{\frac{3}{2}}\)[/tex]
2. [tex]\(\frac{x}{\sqrt{x}} = x^{1 - \frac{1}{2}} = x^{\frac{1}{2}}\)[/tex]
3. [tex]\(\frac{1}{\sqrt{x}} = x^{0 - \frac{1}{2}} = x^{-\frac{1}{2}}\)[/tex]
Substituting these back into the integral, we get:
[tex]\[
\int \left( x^{\frac{3}{2}} + x^{\frac{1}{2}} + x^{-\frac{1}{2}} \right) \, dx
\][/tex]
Now, we can integrate each term separately using the power rule for integration, which states that [tex]\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\)[/tex], where [tex]\(C\)[/tex] is the constant of integration:
1. [tex]\(\int x^{\frac{3}{2}} \, dx = \frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} = \frac{x^{\frac{5}{2}}}{\frac{5}{2}} = \frac{2}{5} x^{\frac{5}{2}}\)[/tex]
2. [tex]\(\int x^{\frac{1}{2}} \, dx = \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3} x^{\frac{3}{2}}\)[/tex]
3. [tex]\(\int x^{-\frac{1}{2}} \, dx = \frac{x^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2x^{\frac{1}{2}}\)[/tex]
Combining these results, the indefinite integral is:
[tex]\[
\int \frac{x^2 + x + 1}{\sqrt{x}} \, dx = \frac{2}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}} + 2x^{\frac{1}{2}} + C
\][/tex]
This simplifies to:
[tex]\[
2 \cdot \frac{1}{5} x^{\frac{5}{2}} + 2 \cdot \frac{1}{3} x^{\frac{3}{2}} + 2 \cdot 1 x^{\frac{1}{2}} + C
\][/tex]
[tex]\[
= \frac{2}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}} + 2 \sqrt{x} + C
\][/tex]
Thus, the final answer is:
[tex]\[
\boxed{\frac{2}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}} + 2 \sqrt{x} + C}
\][/tex]
