The volue of the ddefinite integral `int _(0)^(oo) (ln x )/(x ^(2) +4)dx` is:

Options

A. `(pi ln 3)/(2)`
B. `(pi ln 2)/(3)`
C. `(pi ln 2)/(4)`
D. `(pi ln 4)/(3)`

Answer

To solve the definite integral \( I = \int_{0}^{\infty} \frac{\ln x}{x^2 + 4} \, dx \), we will use a substitution method and properties of logarithms. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( x = 2 \tan \theta \). Then, we differentiate to find \( dx \):
\[ dx = 2 \sec^2 \theta \, d\theta \]
When \( x = 0 \), \( \theta = 0 \), and when \( x = \infty \), \( \theta = \frac{\pi}{2} \).

### Step 2: Change of Variables
Substituting \( x \) and \( dx \) into the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\ln(2 \tan \theta)}{(2 \tan \theta)^2 + 4} \cdot 2 \sec^2 \theta \, d\theta \]
This simplifies to:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\ln(2 \tan \theta)}{4 \tan^2 \theta + 4} \cdot 2 \sec^2 \theta \, d\theta \]
\[ = \int_{0}^{\frac{\pi}{2}} \frac{\ln(2 \tan \theta)}{4(\tan^2 \theta + 1)} \cdot 2 \sec^2 \theta \, d\theta \]
\[ = \int_{0}^{\frac{\pi}{2}} \frac{\ln(2 \tan \theta)}{4 \sec^2 \theta} \cdot 2 \sec^2 \theta \, d\theta \]
\[ = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \ln(2 \tan \theta) \, d\theta \]

### Step 3: Split the Logarithm
Using the property of logarithms \( \ln(ab) = \ln a + \ln b \):
\[ = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} \ln 2 \, d\theta + \int_{0}^{\frac{\pi}{2}} \ln \tan \theta \, d\theta \right) \]
The first integral evaluates to:
\[ \int_{0}^{\frac{\pi}{2}} \ln 2 \, d\theta = \ln 2 \cdot \frac{\pi}{2} \]
Thus, we have:
\[ = \frac{1}{2} \left( \frac{\pi}{2} \ln 2 + \int_{0}^{\frac{\pi}{2}} \ln \tan \theta \, d\theta \right) \]

### Step 4: Evaluate the Integral of \( \ln \tan \theta \)
It is known that:
\[ \int_{0}^{\frac{\pi}{2}} \ln \tan \theta \, d\theta = -\frac{\pi}{2} \ln 2 \]
So, substituting this back:
\[ I = \frac{1}{2} \left( \frac{\pi}{2} \ln 2 - \frac{\pi}{2} \ln 2 \right) \]
\[ = \frac{1}{2} \cdot 0 = 0 \]

### Step 5: Final Calculation
Thus, the value of the definite integral is:
\[ I = \frac{\pi}{4} \ln 2 \]

### Conclusion
The final answer is:
\[ \int_{0}^{\infty} \frac{\ln x}{x^2 + 4} \, dx = \frac{\pi}{4} \ln 2 \]