The given integral is:

$\int \frac{x^2 - 3x + 2}{\sqrt{x}} \, dx$

Solution:

We can simplify the integrand by dividing each term in the numerator by $\sqrt{x}$. Recall that $\frac{x^n}{x^m} = x^{n-m}$, so:

$\frac{x^2}{\sqrt{x}} = x^{2 - \frac{1}{2}} = x^{\frac{3}{2}}, \quad \frac{3x}{\sqrt{x}} = 3x^{1 - \frac{1}{2}} = 3x^{\frac{1}{2}}, \quad \frac{2}{\sqrt{x}} = 2x^{-\frac{1}{2}}.$

Thus, the integrand becomes:

$x^{\frac{3}{2}} - 3x^{\frac{1}{2}} + 2x^{-\frac{1}{2}}.$

The integral can now be written as:

$\int \left( x^{\frac{3}{2}} - 3x^{\frac{1}{2}} + 2x^{-\frac{1}{2}} \right) dx.$

Step-by-step Integration:

1. Integrate $x^{\frac{3}{2}}$: $\int x^{\frac{3}{2}} \, dx = \frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} = \frac{x^{\frac{5}{2}}}{\frac{5}{2}} = \frac{2}{5} x^{\frac{5}{2}}.$

2. Integrate $-3x^{\frac{1}{2}}$: $\int -3x^{\frac{1}{2}} \, dx = -3 \int x^{\frac{1}{2}} \, dx = -3 \cdot \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} = -3 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = -2x^{\frac{3}{2}}.$

3. Integrate $2x^{-\frac{1}{2}}$: $\int 2x^{-\frac{1}{2}} \, dx = 2 \int x^{-\frac{1}{2}} \, dx = 2 \cdot \frac{x^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} = 2 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 4x^{\frac{1}{2}}.$

Combine Results:

Now, summing all terms together:

$\int \frac{x^2 - 3x + 2}{\sqrt{x}} \, dx = \frac{2}{5}x^{\frac{5}{2}} - 2x^{\frac{3}{2}} + 4x^{\frac{1}{2}} + C,$

where $C$ is the constant of integration.

Would you like me to go into more detail or provide related examples?

5 Related Questions:

1. How do you solve integrals involving polynomials divided by roots of $x$?
2. Can we apply substitution to solve this integral instead of simplifying directly?
3. How do you handle definite integrals for similar functions?
4. What is the geometric interpretation of the result of this integral?
5. How do fractional exponents work when integrating or differentiating?

Tip:

Always simplify the integrand before attempting integration—it often makes the process significantly easier.