`int _(3)^(8) (2-3x)/(xsqrt(1+x))dx` is equal to

Options

A. `2log((3)/(2e^(3)))`
B. `log((3)/(e^(3)))`
C. `4log((3)/(e^(3)))`
D. None of these

Explanation

To solve the integral \(\int_{3}^{8} \frac{2 - 3x}{x \sqrt{1+x}} \, dx\), we will perform a substitution and then evaluate the integral step by step.

### Step 1: Substitution
Let \( t^2 = 1 + x \). Then, we have:
\[ x = t^2 - 1 \]
Differentiating both sides gives:
\[ dx = 2t \, dt \]

### Step 2: Change the limits of integration
When \( x = 3 \):
\[ t^2 = 1 + 3 = 4 \implies t = 2 \]
When \( x = 8 \):
\[ t^2 = 1 + 8 = 9 \implies t = 3 \]
Thus, the limits change from \( x = 3 \) to \( x = 8 \) into \( t = 2 \) to \( t = 3 \).

### Step 3: Substitute into the integral
Now, substituting \( x \) and \( dx \) into the integral:
\[ \int_{3}^{8} \frac{2 - 3x}{x \sqrt{1+x}} \, dx = \int_{2}^{3} \frac{2 - 3(t^2 - 1)}{(t^2 - 1) \sqrt{t^2}} \cdot 2t \, dt \]
This simplifies to:
\[ = \int_{2}^{3} \frac{2 - 3t^2 + 3}{(t^2 - 1)t} \cdot 2t \, dt = \int_{2}^{3} \frac{5 - 3t^2}{t^2 - 1} \cdot 2 \, dt \]

### Step 4: Simplify the integral
Now we can separate the integral:
\[ = 2 \int_{2}^{3} \frac{5 - 3t^2}{t^2 - 1} \, dt = 2 \left( \int_{2}^{3} \frac{5}{t^2 - 1} \, dt - 3 \int_{2}^{3} dt \right) \]

### Step 5: Evaluate the integrals
1. **First Integral**:
\[ \int \frac{5}{t^2 - 1} \, dt = 5 \cdot \frac{1}{2} \ln \left| \frac{t-1}{t+1} \right| + C = \frac{5}{2} \ln \left| \frac{t-1}{t+1} \right| + C \]
Evaluating from 2 to 3:
\[ = \frac{5}{2} \left( \ln \left| \frac{3-1}{3+1} \right| - \ln \left| \frac{2-1}{2+1} \right| \right) = \frac{5}{2} \left( \ln \left( \frac{2}{4} \right) - \ln \left( \frac{1}{3} \right) \right) = \frac{5}{2} \left( \ln \left( \frac{2}{4} \cdot 3 \right) \right) = \frac{5}{2} \ln \left( \frac{3}{2} \right) \]

2. **Second Integral**:
\[ \int dt = t \bigg|_{2}^{3} = 3 - 2 = 1 \]

### Step 6: Combine the results
Putting it all together:
\[ = 2 \left( \frac{5}{2} \ln \left( \frac{3}{2} \right) - 3 \cdot 1 \right) = 5 \ln \left( \frac{3}{2} \right) - 6 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{3}^{8} \frac{2 - 3x}{x \sqrt{1+x}} \, dx = 5 \ln \left( \frac{3}{2} \right) - 6 \]