The value of `int_(3)^(5) (x^(2))/(x^(2)-4)`dx, is

Options

A. `2-log_(e )((15)/(7))`
B. `2+log_(e )((15)/(7))`
C. `2+4log_(e )3-4 log_(e )7+4 log_(e )5`
D. `2-tan^(-1)((15)/(7))`

Answer

To solve the integral \( I = \int_{3}^{5} \frac{x^2}{x^2 - 4} \, dx \), we can break it down into simpler parts. Here are the steps:

### Step 1: Rewrite the integrand
We can rewrite the integrand by splitting the fraction:
\[ \frac{x^2}{x^2 - 4} = \frac{x^2 - 4 + 4}{x^2 - 4} = \frac{x^2 - 4}{x^2 - 4} + \frac{4}{x^2 - 4} = 1 + \frac{4}{x^2 - 4} \]
Thus, we have:
\[ I = \int_{3}^{5} \left( 1 + \frac{4}{x^2 - 4} \right) \, dx \]

### Step 2: Separate the integral
Now we can separate the integral:
\[ I = \int_{3}^{5} 1 \, dx + \int_{3}^{5} \frac{4}{x^2 - 4} \, dx \]

### Step 3: Evaluate the first integral
The first integral is straightforward:
\[ \int_{3}^{5} 1 \, dx = [x]_{3}^{5} = 5 - 3 = 2 \]

### Step 4: Evaluate the second integral
For the second integral, we can use the formula for the integral of \( \frac{1}{x^2 - a^2} \):
\[ \int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| + C \]
In our case, \( a = 2 \), so:
\[ \int \frac{4}{x^2 - 4} \, dx = 4 \cdot \frac{1}{2 \cdot 2} \ln \left| \frac{x - 2}{x + 2} \right| = \ln \left| \frac{x - 2}{x + 2} \right| \]
Thus,
\[ \int_{3}^{5} \frac{4}{x^2 - 4} \, dx = \left[ \ln \left| \frac{x - 2}{x + 2} \right| \right]_{3}^{5} \]

### Step 5: Calculate the limits for the second integral
Now we evaluate the limits:
\[ \left[ \ln \left| \frac{x - 2}{x + 2} \right| \right]_{3}^{5} = \ln \left| \frac{5 - 2}{5 + 2} \right| - \ln \left| \frac{3 - 2}{3 + 2} \right| \]
Calculating these:
\[ = \ln \left| \frac{3}{7} \right| - \ln \left| \frac{1}{5} \right| = \ln \left( \frac{3}{7} \cdot 5 \right) = \ln \left( \frac{15}{7} \right) \]

### Step 6: Combine the results
Now we can combine the results:
\[ I = 2 + \ln \left( \frac{15}{7} \right) \]

### Final Result
Thus, the value of the integral is:
\[ I = 2 + \ln \left( \frac{15}{7} \right) \]