`int _(-2) ^(3) (1)/(x +5) dx =`

Options

A. `log ((3)/( 8)) `
B. `log ((8)/(3))`
C. `log ((8)/(5))`
D. 0

Answer

To solve the integral \(\int_{-2}^{3} \frac{1}{x + 5} \, dx\), we can follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{-2}^{3} \frac{1}{x + 5} \, dx \]

### Step 2: Use substitution
Let \(t = x + 5\). Then, we differentiate:
\[ \frac{dt}{dx} = 1 \implies dx = dt \]

### Step 3: Change the limits of integration
Now we need to change the limits of integration according to our substitution:
- When \(x = -2\), \(t = -2 + 5 = 3\)
- When \(x = 3\), \(t = 3 + 5 = 8\)

So, the integral becomes:
\[ I = \int_{3}^{8} \frac{1}{t} \, dt \]

### Step 4: Integrate
The integral of \(\frac{1}{t}\) is \(\log |t|\). Therefore, we have:
\[ I = \left[ \log |t| \right]_{3}^{8} \]

### Step 5: Evaluate the definite integral
Now we evaluate this from 3 to 8:
\[ I = \log |8| - \log |3| = \log 8 - \log 3 \]

### Step 6: Simplify using logarithmic properties
Using the property of logarithms that states \(\log a - \log b = \log \frac{a}{b}\):
\[ I = \log \frac{8}{3} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-2}^{3} \frac{1}{x + 5} \, dx = \log \frac{8}{3} \]

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