Evaluate `int_(0)^(infty)(lnxdx)/(x^(2)+2x+4)`

Answer

To evaluate the integral

\[ I = \int_{0}^{\infty} \frac{\ln x}{x^2 + 2x + 4} \, dx, \]

we will follow a systematic approach.

### Step 1: Rewrite the Integral
We start by rewriting the denominator:

\[ x^2 + 2x + 4 = (x+1)^2 + 3. \]

Thus, we can express the integral as:

\[ I = \int_{0}^{\infty} \frac{\ln x}{(x+1)^2 + 3} \, dx. \]

### Step 2: Use a Substitution
Next, we will use the substitution \( x = 2t \). Then, \( dx = 2 \, dt \) and the limits remain the same:

\[ I = \int_{0}^{\infty} \frac{\ln(2t)}{(2t)^2 + 2(2t) + 4} \cdot 2 \, dt. \]

This simplifies to:

\[ I = 2 \int_{0}^{\infty} \frac{\ln(2t)}{4t^2 + 8t + 4} \, dt = 2 \int_{0}^{\infty} \frac{\ln(2t)}{4(t^2 + 2t + 1)} \, dt. \]

### Step 3: Simplify Further
We can simplify the denominator:

\[ 4(t^2 + 2t + 1) = 4(t + 1)^2. \]

Thus, we have:

\[ I = \frac{1}{2} \int_{0}^{\infty} \frac{\ln(2t)}{(t + 1)^2} \, dt. \]

### Step 4: Split the Logarithm
Using the property of logarithms, we can split the integral:

\[ I = \frac{1}{2} \left( \int_{0}^{\infty} \frac{\ln 2}{(t + 1)^2} \, dt + \int_{0}^{\infty} \frac{\ln t}{(t + 1)^2} \, dt \right). \]

### Step 5: Evaluate the First Integral
The first integral can be evaluated:

\[ \int_{0}^{\infty} \frac{\ln 2}{(t + 1)^2} \, dt = \ln 2 \cdot \int_{0}^{\infty} \frac{1}{(t + 1)^2} \, dt = \ln 2 \cdot 1 = \ln 2. \]

### Step 6: Evaluate the Second Integral
The second integral can be evaluated using the known result:

\[ \int_{0}^{\infty} \frac{\ln t}{(t + 1)^2} \, dt = -\frac{\pi^2}{6}. \]

### Step 7: Combine Results
Now we can combine the results:

\[ I = \frac{1}{2} \left( \ln 2 - \frac{\pi^2}{6} \right). \]

### Final Answer
Thus, the final answer for the integral is:

\[ I = \frac{1}{2} \ln 2 - \frac{\pi^2}{12}. \]