If `2f(x) + 3f((1)/(x))= (1)/(x)-2, x ne 0` then `int_(1)^(2) f(x)dx`=

Options

A. `-(2)/(5) log 2 + (1)/(2)`
B. `-(2)/(5) log 2- (1)/(2)`
C. `(2)/(5) log 2+ (1)/(2)`
D. none

Answer

To solve the problem, we need to find the value of the integral \( \int_{1}^{2} f(x) \, dx \) given the equation:

\[ 2f(x) + 3f\left(\frac{1}{x}\right) = \frac{1}{x} - 2 \]

### Step 1: Substitute \( x \) with \( \frac{1}{x} \)

We will replace \( x \) with \( \frac{1}{x} \) in the original equation:

\[ 2f\left(\frac{1}{x}\right) + 3f(x) = x - 2 \]

### Step 2: Write the two equations

Now we have two equations:

1. \( 2f(x) + 3f\left(\frac{1}{x}\right) = \frac{1}{x} - 2 \) (Equation 1)
2. \( 2f\left(\frac{1}{x}\right) + 3f(x) = x - 2 \) (Equation 2)

### Step 3: Multiply Equation 1 by 2 and Equation 2 by 3

Multiply Equation 1 by 2:

\[ 4f(x) + 6f\left(\frac{1}{x}\right) = \frac{2}{x} - 4 \]

Multiply Equation 2 by 3:

\[ 6f\left(\frac{1}{x}\right) + 9f(x) = 3x - 6 \]

### Step 4: Subtract the two equations

Now, we will subtract the modified Equation 1 from the modified Equation 2:

\[ (6f\left(\frac{1}{x}\right) + 9f(x)) - (4f(x) + 6f\left(\frac{1}{x}\right)) = (3x - 6) - \left(\frac{2}{x} - 4\right) \]

This simplifies to:

\[ 3f(x) = 3x - 6 + 4 - \frac{2}{x} \]

\[ 3f(x) = 3x - 2 - \frac{2}{x} \]

### Step 5: Solve for \( f(x) \)

Now, divide both sides by 3:

\[ f(x) = x - \frac{2}{3} - \frac{2}{3x} \]

### Step 6: Evaluate the integral \( \int_{1}^{2} f(x) \, dx \)

Now we can evaluate the integral:

\[ \int_{1}^{2} f(x) \, dx = \int_{1}^{2} \left( x - \frac{2}{3} - \frac{2}{3x} \right) \, dx \]

### Step 7: Break down the integral

This can be broken down into three separate integrals:

\[ \int_{1}^{2} x \, dx - \frac{2}{3} \int_{1}^{2} 1 \, dx - \frac{2}{3} \int_{1}^{2} \frac{1}{x} \, dx \]

### Step 8: Calculate each integral

1. \( \int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \)
2. \( \int_{1}^{2} 1 \, dx = \left[ x \right]_{1}^{2} = 2 - 1 = 1 \)
3. \( \int_{1}^{2} \frac{1}{x} \, dx = \left[ \log x \right]_{1}^{2} = \log 2 - \log 1 = \log 2 \)

### Step 9: Combine the results

Putting it all together:

\[ \int_{1}^{2} f(x) \, dx = \frac{3}{2} - \frac{2}{3} \cdot 1 - \frac{2}{3} \cdot \log 2 \]

Calculating \( -\frac{2}{3} \):

\[ = \frac{3}{2} - \frac{2}{3} - \frac{2}{3} \log 2 \]

### Step 10: Find a common denominator

The common denominator for \( \frac{3}{2} \) and \( \frac{2}{3} \) is 6:

\[ \frac{3}{2} = \frac{9}{6}, \quad \frac{2}{3} = \frac{4}{6} \]

Thus,

\[ \int_{1}^{2} f(x) \, dx = \frac{9}{6} - \frac{4}{6} - \frac{2}{3} \log 2 = \frac{5}{6} - \frac{2}{3} \log 2 \]

### Final Result

Thus, the value of the integral \( \int_{1}^{2} f(x) \, dx \) is:

\[ \int_{1}^{2} f(x) \, dx = \frac{5}{6} - \frac{2}{3} \log 2 \]