To solve the definite integral \(\int_4^9\left(\frac{3}{\sqrt{x}}-\frac{1}{x \sqrt{x}}\right) dx\), let's break it down step-by-step:
1. Rewrite the integrand:
The given integrand is \(\frac{3}{\sqrt{x}} - \frac{1}{x \sqrt{x}}\). Let's rewrite each term in a simpler form:
[tex]\[
\frac{3}{\sqrt{x}} = 3x^{-\frac{1}{2}}
\][/tex]
[tex]\[
\frac{1}{x \sqrt{x}} = \frac{1}{x \cdot x^{\frac{1}{2}}} = \frac{1}{x^{\frac{3}{2}}} = x^{-\frac{3}{2}}
\][/tex]
Therefore, the integrand becomes:
[tex]\[
3x^{-\frac{1}{2}} - x^{-\frac{3}{2}}
\][/tex]
2. Integrate each term separately:
We now need to find the antiderivative of each term:
[tex]\[
\int 3x^{-\frac{1}{2}} \, dx \quad \text{and} \quad \int x^{-\frac{3}{2}} \, dx
\][/tex]
For the first term:
[tex]\[
\int 3x^{-\frac{1}{2}} \, dx = 3 \int x^{-\frac{1}{2}} \, dx = 3 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 3 \cdot 2x^{\frac{1}{2}} = 6x^{\frac{1}{2}}
\][/tex]
For the second term:
[tex]\[
\int x^{-\frac{3}{2}} \, dx = \int x^{-\frac{3}{2}} \, dx = \frac{x^{-\frac{3}{2}+1}}{-\frac{3}{2}+1} = \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} = -2x^{-\frac{1}{2}}
\][/tex]
3. Combine the antiderivatives:
Combining these results, the antiderivative of the integrand is:
[tex]\[
6x^{\frac{1}{2}} - 2x^{-\frac{1}{2}}
\][/tex]
4. Evaluate the definite integral:
We now need to evaluate the antiderivative at the upper and lower limits (9 and 4):
[tex]\[
\left[ 6x^{\frac{1}{2}} - 2x^{-\frac{1}{2}} \right]_4^9
\][/tex]
Calculate at \(x = 9\):
[tex]\[
6 \cdot 9^{\frac{1}{2}} - 2 \cdot 9^{-\frac{1}{2}} = 6 \cdot 3 - 2 \cdot \frac{1}{3} = 18 - \frac{2}{3} = 18 - 0.66666666666667 = 17.33333333333333
\][/tex]
Calculate at \(x = 4\):
[tex]\[
6 \cdot 4^{\frac{1}{2}} - 2 \cdot 4^{-\frac{1}{2}} = 6 \cdot 2 - 2 \cdot \frac{1}{2} = 12 - 1 = 11
\][/tex]
Taking the difference:
[tex]\[
17.33333333333333 - 11 = 6.33333333333333
\][/tex]
Therefore, the value of the definite integral \(\int_4^9\left(\frac{3}{\sqrt{x}}-\frac{1}{x \sqrt{x}}\right) dx\) is:
[tex]\[
5.66666666666667
\][/tex]
