$\int_{10}^{100}\frac{log_{e}x}{log_{e}x+log_{e}(110-x)}dx=$

Question

Accepted Answer

To solve the integral $\int_{10}^{100}\frac{\log_{e}x}{\log_{e}x+\log_{e}(110-x)}dx$, we can use a symmetry property of definite integrals. The key observation here is that the integrand has a specific symmetry that can be exploited.

### Explanation:

1. **Symmetry of the Integrand:**

Consider the substitution $ x = 110 - t $. Then $ dx = -dt $. The limits of integration change as follows: when $ x = 10 $, $ t = 100 $, and when $ x = 100 $, $ t = 10 $. The integral becomes:

$$
   \int_{100}^{10} \frac{\log_{e}(110-t)}{\log_{e}(110-t) + \log_{e}t} (-dt) = \int_{10}^{100} \frac{\log_{e}(110-t)}{\log_{e}(110-t) + \log_{e}t} dt
   $$

Notice that the form of the integrand is preserved under this substitution, meaning:

$$
   \frac{\log_{e}x}{\log_{e}x + \log_{e}(110-x)} = \frac{\log_{e}(110-x)}{\log_{e}(110-x) + \log_{e}x}
   $$

Therefore, the original integral can be split into two parts:

$$
   \int_{10}^{100} \frac{\log_{e}x}{\log_{e}x + \log_{e}(110-x)} dx + \int_{10}^{100} \frac{\log_{e}(110-x)}{\log_{e}x + \log_{e}(110-x)} dx
   $$

Adding these two integrals gives:

$$
   \int_{10}^{100} 1 \, dx = 100 - 10 = 90
   $$

2. **Conclusion:**

Since the sum of the two integrals is 90, and each integral is symmetric, each must be half of 90. Therefore, the value of the original integral is:

$$
   \frac{90}{2} = 45
   $$

Thus, the correct answer is Option 3: 45.

### Why Other Options Are Incorrect:

- **Option 1 (90):** This would imply that the entire integral evaluates to the sum of the two symmetric parts without considering their equal contribution, which is incorrect.

- **Option 2 (0):** This would suggest that the integral evaluates to zero, which is not possible given the symmetry and positive nature of the integrand over the interval.

- **Option 4 (110):** This would imply an incorrect evaluation of the integral, as it does not account for the symmetry and equal division of the integral into two parts.

Therefore, the correct answer is indeed Option 3: 45, as justified by the symmetry and evaluation of the integral.