If `Isum_(k=1)^(98)int_k^(k+1)(k+1)/(x(x+1))dx ,t h e n:` (a) `I<(49)/(50)` (b) `I >(log)_e 99` (c) `I >(49)/(50)` (d) `I<(log)_e 99`

Options

A. `Igtlog_(e)99`
B. `Iltlog_(e)99`
C. `I lt 49/50`
D. `I gt 49/50`

Explanation

To solve the given problem, we need to evaluate the integral

\[ I = \sum_{k=1}^{98} \int_k^{k+1} \frac{k+1}{x(x+1)} \, dx \]

### Step 1: Evaluate the Integral

We start by evaluating the integral

\[ \int_k^{k+1} \frac{k+1}{x(x+1)} \, dx \]

We can simplify the integrand using partial fractions:

\[ \frac{k+1}{x(x+1)} = \frac{k+1}{x} - \frac{k+1}{x+1} \]

Thus, we can rewrite the integral as:

\[ \int_k^{k+1} \left( \frac{k+1}{x} - \frac{k+1}{x+1} \right) \, dx \]

### Step 2: Integrate Each Term

Now we can integrate each term separately:

1. For the first term:

\[ \int_k^{k+1} \frac{k+1}{x} \, dx = (k+1) \left[ \ln x \right]_k^{k+1} = (k+1) (\ln(k+1) - \ln k) = (k+1) \ln\left(\frac{k+1}{k}\right) \]

2. For the second term:

\[ \int_k^{k+1} \frac{k+1}{x+1} \, dx = (k+1) \left[ \ln(x+1) \right]_k^{k+1} = (k+1) (\ln(k+2) - \ln(k+1)) = (k+1) \ln\left(\frac{k+2}{k+1}\right) \]

### Step 3: Combine the Results

Combining both results, we have:

\[ \int_k^{k+1} \frac{k+1}{x(x+1)} \, dx = (k+1) \ln\left(\frac{k+1}{k}\right) - (k+1) \ln\left(\frac{k+2}{k+1}\right) \]

This simplifies to:

\[ (k+1) \left( \ln\left(\frac{k+1}{k}\right) - \ln\left(\frac{k+2}{k+1}\right) \right) = (k+1) \ln\left(\frac{k+1}{k+2}\right) \]

### Step 4: Sum Over k

Now we substitute this back into the summation:

\[ I = \sum_{k=1}^{98} (k+1) \ln\left(\frac{k+1}{k+2}\right) \]

### Step 5: Simplify the Summation

This can be rewritten as:

\[ I = \sum_{k=1}^{98} (k+1) \left( \ln(k+1) - \ln(k+2) \right) \]

### Step 6: Telescoping Series

This summation is telescoping. The terms will cancel out, leading to:

\[ I = \ln(2) + \ln(3) + \ldots + \ln(99) - \ln(3) - \ln(4) - \ldots - \ln(100) \]

After cancellation, we find:

\[ I = \ln\left(\frac{99}{2}\right) \]

### Step 7: Analyze the Result

Now we need to analyze the bounds for \( I \):

1. Since \( \ln(50) < \ln\left(\frac{99}{2}\right) < \ln(99) \), we can conclude:

\[ \ln(50) < I < \ln(99) \]

### Conclusion

Thus, the correct options are:

- \( I > \frac{49}{50} \) (Option c)
- \( I < \ln(99) \) (Option d)