Let `I_(1)=int_(0)^(1)(|lnx|)/(x^(2)+4x+1)dx` and `I_(2)=int_(1)^(oo)(lnx)/(x^(2)+4x+1)dx`, then

Options

A. `I_(1)=I_(2)`
B. `I_(1) gt I_(2)`
C. `I_(1)+I_(2)=0`
D. `I_(1)=2I_(2)`

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and find the relationship between them.

### Step 1: Evaluate \( I_1 \)

Given:
\[ I_1 = \int_0^1 \frac{|\ln x|}{x^2 + 4x + 1} \, dx \]

Since \( \ln x < 0 \) for \( x \in (0, 1) \), we can replace \( |\ln x| \) with \( -\ln x \):
\[ I_1 = \int_0^1 \frac{-\ln x}{x^2 + 4x + 1} \, dx \]

### Step 2: Substitute \( x = \frac{1}{t} \)

Let's use the substitution \( x = \frac{1}{t} \). Then, \( dx = -\frac{1}{t^2} \, dt \). The limits change as follows:
- When \( x = 0 \), \( t \to \infty \)
- When \( x = 1 \), \( t = 1 \)

Thus, we have:
\[ I_1 = \int_{\infty}^1 \frac{-\ln\left(\frac{1}{t}\right)}{\left(\frac{1}{t}\right)^2 + 4\left(\frac{1}{t}\right) + 1} \left(-\frac{1}{t^2}\right) dt \]

This simplifies to:
\[ I_1 = \int_1^{\infty} \frac{\ln t}{\frac{1}{t^2} + \frac{4}{t} + 1} \cdot \frac{1}{t^2} \, dt \]

### Step 3: Simplify the Denominator

The denominator becomes:
\[ \frac{1}{t^2} + \frac{4}{t} + 1 = \frac{1 + 4t + t^2}{t^2} \]
Thus, we can rewrite \( I_1 \):
\[ I_1 = \int_1^{\infty} \frac{\ln t}{\frac{1 + 4t + t^2}{t^2}} \cdot \frac{1}{t^2} \, dt = \int_1^{\infty} \frac{\ln t \cdot t^2}{1 + 4t + t^2} \cdot \frac{1}{t^2} \, dt \]
This simplifies to:
\[ I_1 = \int_1^{\infty} \frac{\ln t}{1 + 4t + t^2} \, dt \]

### Step 4: Recognize \( I_2 \)

Now, we recognize that:
\[ I_2 = \int_1^{\infty} \frac{\ln x}{x^2 + 4x + 1} \, dx \]

### Step 5: Conclude the Relationship

From the previous steps, we have shown that:
\[ I_1 = I_2 \]

Thus, the final result is:
\[ I_1 = I_2 \]