Let `I_(1)=int_(1)^(2)(x)/(sqrt(1+x^(2)))dx` and `I_(2)=int_(1)^(2)(1)/(x)dx`.Then

Options

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

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and then compare their values.

### Step 1: Evaluate \( I_1 = \int_{1}^{2} \frac{x}{\sqrt{1+x^2}} \, dx \)

1. **Substitution**: Let \( x^2 = t \). Then, \( 2x \, dx = dt \) or \( x \, dx = \frac{dt}{2} \).
2. **Change of limits**: When \( x = 1 \), \( t = 1^2 = 1 \). When \( x = 2 \), \( t = 2^2 = 4 \).
3. **Rewrite the integral**:
\[ I_1 = \int_{1}^{4} \frac{\frac{dt}{2}}{\sqrt{1+t}} = \frac{1}{2} \int_{1}^{4} \frac{dt}{\sqrt{1+t}} \]
4. **Further substitution**: Let \( 1+t = y \), then \( dt = dy \).
5. **Change of limits**: When \( t = 1 \), \( y = 2 \). When \( t = 4 \), \( y = 5 \).
6. **Rewrite the integral**:
\[ I_1 = \frac{1}{2} \int_{2}^{5} \frac{dy}{\sqrt{y}} = \frac{1}{2} \int_{2}^{5} y^{-1/2} \, dy \]
7. **Integrate**:
\[ I_1 = \frac{1}{2} \left[ 2y^{1/2} \right]_{2}^{5} = \left[ y^{1/2} \right]_{2}^{5} = \sqrt{5} - \sqrt{2} \]

### Step 2: Evaluate \( I_2 = \int_{1}^{2} \frac{1}{x} \, dx \)

1. **Integrate**: The integral of \( \frac{1}{x} \) is \( \log x \).
2. **Evaluate the limits**:
\[ I_2 = \left[ \log x \right]_{1}^{2} = \log 2 - \log 1 = \log 2 \]
Since \( \log 1 = 0 \), we have \( I_2 = \log 2 \).

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

1. **Approximate values**:
- \( I_1 = \sqrt{5} - \sqrt{2} \)
- \( I_2 = \log 2 \)

2. **Numerical values**:
- \( \sqrt{5} \approx 2.236 \)
- \( \sqrt{2} \approx 1.414 \)
- Thus, \( I_1 \approx 2.236 - 1.414 = 0.822 \)
- \( \log 2 \approx 0.301 \)

3. **Conclusion**: Since \( 0.822 > 0.301 \), we conclude that \( I_1 > I_2 \).

### Final Result
Thus, we find that \( I_1 > I_2 \).

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