If `a, b, gt 0`, then `int_(1)^(a)(1)/(x)dx + int_(1)^(b)(1)/(x)dx=`

Options

A. `int_(1)^(a+1)(1)/(x)dx`
B. `int_(a)^(b)(1)/(x)dx`
C. `int_(1)^(ab)(1)/(x)dx`
D. `int_(1//a)^(1//b)x dx`

Answer

To solve the given problem, we need to evaluate the following expression:

\[ \int_{1}^{a} \frac{1}{x} \, dx + \int_{1}^{b} \frac{1}{x} \, dx \]

### Step 1: Evaluate the first integral

The first integral is:

\[ \int_{1}^{a} \frac{1}{x} \, dx \]

Using the formula for the integral of \( \frac{1}{x} \), we have:

\[ \int \frac{1}{x} \, dx = \log x + C \]

So, we evaluate the definite integral:

\[ \int_{1}^{a} \frac{1}{x} \, dx = \left[ \log x \right]_{1}^{a} = \log a - \log 1 \]

Since \( \log 1 = 0 \):

\[ \int_{1}^{a} \frac{1}{x} \, dx = \log a \]

### Step 2: Evaluate the second integral

Now, we evaluate the second integral:

\[ \int_{1}^{b} \frac{1}{x} \, dx \]

Using the same formula:

\[ \int_{1}^{b} \frac{1}{x} \, dx = \left[ \log x \right]_{1}^{b} = \log b - \log 1 \]

Again, since \( \log 1 = 0 \):

\[ \int_{1}^{b} \frac{1}{x} \, dx = \log b \]

### Step 3: Combine the results

Now we combine the results of both integrals:

\[ \int_{1}^{a} \frac{1}{x} \, dx + \int_{1}^{b} \frac{1}{x} \, dx = \log a + \log b \]

Using the properties of logarithms, we can simplify this:

\[ \log a + \log b = \log(ab) \]

### Final Result

Thus, the final result is:

\[ \int_{1}^{a} \frac{1}{x} \, dx + \int_{1}^{b} \frac{1}{x} \, dx = \log(ab) \]

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