To solve the integral [tex]\(\int \left[ \frac{(1-x) \sqrt{x}}{x} \right] \, dx\)[/tex], we can proceed with the following steps:
1. Simplify the Integrand:
The given integrand is [tex]\(\frac{(1 - x) \sqrt{x}}{x}\)[/tex]. We can split the fraction:
[tex]\[
\frac{(1 - x) \sqrt{x}}{x} = \frac{1 \cdot \sqrt{x}}{x} - \frac{x \cdot \sqrt{x}}{x}
\][/tex]
Simplify each term separately:
[tex]\[
\frac{\sqrt{x}}{x} = \frac{x^{1/2}}{x} = x^{1/2 - 1} = x^{-1/2}
\][/tex]
[tex]\[
\frac{x \cdot \sqrt{x}}{x} = \sqrt{x} = x^{1/2}
\][/tex]
Thus, the integrand simplifies to:
[tex]\[
x^{-1/2} - x^{1/2}
\][/tex]
2. Integrate the Simplified Expression:
Now, we need to integrate each term [tex]\( x^{-1/2} \)[/tex] and [tex]\( -x^{1/2} \)[/tex] separately.
Recall the power rule for integration:
[tex]\[
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{(for } n \neq -1 \text{)}
\][/tex]
Let's integrate [tex]\( x^{-1/2} \)[/tex]:
[tex]\[
\int x^{-1/2} \, dx = \int x^{-\frac{1}{2}} \, dx = \frac{x^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2x^{1/2} = 2 \sqrt{x}
\][/tex]
Now, integrate [tex]\( -x^{1/2} \)[/tex]:
[tex]\[
\int -x^{1/2} \, dx = -\int x^{1/2} \, dx = -\left( \frac{x^{1/2 + 1}}{1/2 + 1} \right) = -\left( \frac{x^{3/2}}{3/2} \right) = -\left( \frac{2}{3} x^{3/2} \right) = -\frac{2}{3} x^{3/2}
\][/tex]
3. Combine the Results:
The final result of the integral [tex]\(\int (x^{-1/2} - x^{1/2}) \, dx\)[/tex] is given by combining the integrated terms:
[tex]\[
\int \left( x^{-1/2} - x^{1/2} \right) \, dx = 2 \sqrt{x} - \frac{2}{3} x^{3/2} + C
\][/tex]
Hence, we have:
[tex]\[
\int \left[ \frac{(1 - x) \sqrt{x}}{x} \right] \, dx = 2 \sqrt{x} - \frac{2}{3} x^{3/2} + C
\][/tex]
So, the final answer is:
[tex]\[
- \frac{2}{3} x^{3/2} + 2 \sqrt{x} + C
\][/tex]
