Let `f (x)=int(2x)/((x^2+1)(x^2+3))dx`. if `f (3)=1/2(log_e5-log_e6)`,then `f(4)`is equal to

Options

A. `log_e 17-log_e 18`
B. `log_e 19-log_e 20`
C. `1/2(log_e 19-log_e 17)`
D. `1/2(log_e 17-log_e 19)`

Answer

To solve the problem, we start with the function defined as:

\[ f(x) = \int \frac{2x}{(x^2 + 1)(x^2 + 3)} \, dx \]

### Step 1: Substitution
We will use the substitution \( t = x^2 \). Then, we have:

\[ dt = 2x \, dx \implies dx = \frac{dt}{2x} \]

Substituting \( x^2 = t \) gives us:

\[ f(x) = \int \frac{2x}{(t + 1)(t + 3)} \cdot \frac{dt}{2x} = \int \frac{1}{(t + 1)(t + 3)} \, dt \]

### Step 2: Partial Fraction Decomposition
Next, we perform partial fraction decomposition on the integrand:

\[ \frac{1}{(t + 1)(t + 3)} = \frac{A}{t + 1} + \frac{B}{t + 3} \]

Multiplying through by the denominator \((t + 1)(t + 3)\) gives:

\[ 1 = A(t + 3) + B(t + 1) \]

Expanding and rearranging:

\[ 1 = (A + B)t + (3A + B) \]

This leads to the system of equations:

1. \( A + B = 0 \)
2. \( 3A + B = 1 \)

### Step 3: Solving the System of Equations
From the first equation, we have \( B = -A \). Substituting into the second equation:

\[ 3A - A = 1 \implies 2A = 1 \implies A = \frac{1}{2} \]

Thus, \( B = -\frac{1}{2} \). Therefore, we can write:

\[ \frac{1}{(t + 1)(t + 3)} = \frac{1/2}{t + 1} - \frac{1/2}{t + 3} \]

### Step 4: Integrating
Now we can integrate:

\[ f(x) = \int \left( \frac{1/2}{t + 1} - \frac{1/2}{t + 3} \right) dt = \frac{1}{2} \ln |t + 1| - \frac{1}{2} \ln |t + 3| + C \]

Substituting back \( t = x^2 \):

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

### Step 5: Finding the Constant \( C \)
We know that \( f(3) = \frac{1}{2} (\ln 5 - \ln 6) \). Thus:

\[ f(3) = \frac{1}{2} \ln \left( \frac{3^2 + 1}{3^2 + 3} \right) + C = \frac{1}{2} \ln \left( \frac{10}{12} \right) + C = \frac{1}{2} \ln \left( \frac{5}{6} \right) + C \]

Setting this equal to \( \frac{1}{2} (\ln 5 - \ln 6) \):

\[ \frac{1}{2} \ln \left( \frac{5}{6} \right) + C = \frac{1}{2} \ln \left( \frac{5}{6} \right) \]

This implies \( C = 0 \).

### Step 6: Final Function
Thus, we have:

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

### Step 7: Calculate \( f(4) \)
Now we calculate \( f(4) \):

\[ f(4) = \frac{1}{2} \ln \left( \frac{4^2 + 1}{4^2 + 3} \right) = \frac{1}{2} \ln \left( \frac{16 + 1}{16 + 3} \right) = \frac{1}{2} \ln \left( \frac{17}{19} \right) \]

### Final Answer
Thus, the value of \( f(4) \) is:

\[ f(4) = \frac{1}{2} \ln \left( \frac{17}{19} \right) \]