Consider `I_(1)=int_(10)^(20)(lnx)/(lnx+ln(30-x))dx` and `I_(2)=int_(20)^(30)(lnx)/(lnx +ln(50-x))dx`. Then, the value of `(I_(1))/(I_(2))` is

Options

A. 10
B. 2
C. 1
D. `(1)/(2)`

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and then find the ratio \( \frac{I_1}{I_2} \).

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

Given:
\[ I_1 = \int_{10}^{20} \frac{\ln x}{\ln x + \ln(30 - x)} \, dx \]

Using the property of integrals:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
we can substitute \( x \) with \( 30 - x \):
\[ I_1 = \int_{10}^{20} \frac{\ln(30 - x)}{\ln(30 - x) + \ln x} \, dx \]

Now, we can add this expression to the original \( I_1 \):
\[ I_1 + I_1 = \int_{10}^{20} \left( \frac{\ln x}{\ln x + \ln(30 - x)} + \frac{\ln(30 - x)}{\ln(30 - x) + \ln x} \right) dx \]

The sum of the fractions simplifies to:
\[ \frac{\ln x + \ln(30 - x)}{\ln x + \ln(30 - x)} = 1 \]

Thus:
\[ 2I_1 = \int_{10}^{20} 1 \, dx = [x]_{10}^{20} = 20 - 10 = 10 \]

So,
\[ I_1 = \frac{10}{2} = 5 \]

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

Given:
\[ I_2 = \int_{20}^{30} \frac{\ln x}{\ln x + \ln(50 - x)} \, dx \]

Using the same property of integrals:
\[ I_2 = \int_{20}^{30} \frac{\ln(50 - x)}{\ln(50 - x) + \ln x} \, dx \]

Adding this to \( I_2 \):
\[ I_2 + I_2 = \int_{20}^{30} \left( \frac{\ln x}{\ln x + \ln(50 - x)} + \frac{\ln(50 - x)}{\ln(50 - x) + \ln x} \right) dx \]

Again, the sum simplifies to:
\[ \frac{\ln x + \ln(50 - x)}{\ln x + \ln(50 - x)} = 1 \]

Thus:
\[ 2I_2 = \int_{20}^{30} 1 \, dx = [x]_{20}^{30} = 30 - 20 = 10 \]

So,
\[ I_2 = \frac{10}{2} = 5 \]

### Step 3: Calculate \( \frac{I_1}{I_2} \)

Now that we have both integrals:
\[ I_1 = 5 \quad \text{and} \quad I_2 = 5 \]

Thus,
\[ \frac{I_1}{I_2} = \frac{5}{5} = 1 \]

### Final Answer
\[ \frac{I_1}{I_2} = 1 \]