Certainly! Let's solve the integral [tex]\(\int \left[\frac{(1-x) \sqrt{x}}{x}\right] dx\)[/tex] step-by-step.
1. Simplify the integrand:
[tex]\[
\frac{(1-x) \sqrt{x}}{x}
\][/tex]
We can split the fraction inside the integrand:
[tex]\[
\frac{1 \cdot \sqrt{x}}{x} - \frac{x \cdot \sqrt{x}}{x}
\][/tex]
Simplify each term separately:
[tex]\[
\frac{\sqrt{x}}{x} - \sqrt{x}
\][/tex]
Next, rewrite [tex]\(\frac{\sqrt{x}}{x}\)[/tex] as [tex]\(\frac{x^{1/2}}{x} = x^{1/2 - 1} = x^{-1/2}\)[/tex]. Thus, the simplified integrand becomes:
[tex]\[
x^{-1/2} - x^{1/2}
\][/tex]
2. Set up the integral:
[tex]\[
\int \left[x^{-1/2} - x^{1/2}\right] dx
\][/tex]
3. Integrate each term separately:
[tex]\[
\int x^{-1/2} \, dx - \int x^{1/2} \, dx
\][/tex]
4. Find the antiderivative of each term:
- For [tex]\(\int x^{-1/2} \, dx\)[/tex], we use the power rule for integration which states [tex]\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\)[/tex] (where [tex]\(n \neq -1\)[/tex]).
Here, [tex]\(n = -1/2\)[/tex]:
[tex]\[
\int x^{-1/2} \, dx = \frac{x^{(-1/2) + 1}}{(-1/2) + 1} = \frac{x^{1/2}}{1/2} = 2x^{1/2} = 2\sqrt{x}
\][/tex]
- For [tex]\(\int x^{1/2} \, dx\)[/tex], again using the power rule for integration where [tex]\(n = 1/2\)[/tex]:
[tex]\[
\int x^{1/2} \, dx = \frac{x^{(1/2) + 1}}{(1/2) + 1} = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2}
\][/tex]
5. Combine the results:
[tex]\[
\int \left[x^{-1/2} - x^{1/2}\right] dx = 2\sqrt{x} - \frac{2}{3} x^{3/2}
\][/tex]
6. Rewrite in a standard form:
[tex]\[
= -\frac{2}{3} x^{3/2} + 2\sqrt{x}
\][/tex]
Thus, the integral evaluates to:
[tex]\[
\int \left[\frac{(1-x) \sqrt{x}}{x}\right] dx = -\frac{2}{3} x^{3/2} + 2\sqrt{x} + C
\][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
