Let `I _(1) = int _(1) ^(2) (dx )/(sqrt(1 + x ^(2))) and I _(2) = int _(1) ^(2) (dx)/(x)`

Options

A. `I _(1) gt I _(2)`
B. `I _(2) gt I _(1)`
C. `I _(1) = I_(2)`
D. `I _(1) gt 2I _(2)`

Answer

To solve the problem, we need to evaluate the two definite integrals \( I_1 \) and \( I_2 \) given by:

\[ I_1 = \int_{1}^{2} \frac{dx}{\sqrt{1 + x^2}} \]

\[ I_2 = \int_{1}^{2} \frac{dx}{x} \]

### Step 1: Evaluate \( I_1 \)

To evaluate \( I_1 \), we use the formula for the integral of \( \frac{dx}{\sqrt{a^2 + x^2}} \):

\[ \int \frac{dx}{\sqrt{a^2 + x^2}} = \ln(x + \sqrt{x^2 + a^2}) + C \]

In our case, \( a = 1 \). Thus, we have:

\[ I_1 = \int_{1}^{2} \frac{dx}{\sqrt{1 + x^2}} = \left[ \ln(x + \sqrt{x^2 + 1}) \right]_{1}^{2} \]

Calculating the limits:

\[ I_1 = \ln(2 + \sqrt{2^2 + 1}) - \ln(1 + \sqrt{1^2 + 1}) \]

\[ = \ln(2 + \sqrt{5}) - \ln(1 + \sqrt{2}) \]

Using the property of logarithms \( \ln a - \ln b = \ln \left( \frac{a}{b} \right) \):

\[ I_1 = \ln \left( \frac{2 + \sqrt{5}}{1 + \sqrt{2}} \right) \]

### Step 2: Evaluate \( I_2 \)

Now, we evaluate \( I_2 \):

\[ I_2 = \int_{1}^{2} \frac{dx}{x} = \left[ \ln x \right]_{1}^{2} \]

Calculating the limits:

\[ I_2 = \ln(2) - \ln(1) = \ln(2) - 0 = \ln(2) \]

### Step 3: Compare \( I_1 \) and \( I_2 \)

Now we have:

\[ I_1 = \ln \left( \frac{2 + \sqrt{5}}{1 + \sqrt{2}} \right) \]

\[ I_2 = \ln(2) \]

To compare \( I_1 \) and \( I_2 \), we need to check whether:

\[ \frac{2 + \sqrt{5}}{1 + \sqrt{2}} < 2 \]

Cross-multiplying gives:

\[ 2 + \sqrt{5} < 2(1 + \sqrt{2}) \]

Simplifying this:

\[ 2 + \sqrt{5} < 2 + 2\sqrt{2} \]

This simplifies to:

\[ \sqrt{5} < 2\sqrt{2} \]

Squaring both sides (since both sides are positive):

\[ 5 < 8 \quad \text{(true)} \]

Thus, we conclude:

\[ I_1 < I_2 \]

### Final Conclusion

The correct option is that \( I_2 \) is greater than \( I_1 \).