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 find the relationship between the integrals \( I_1 \) and \( I_2 \) defined as:

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

\[ I_2 = \int_1^\infty \frac{\ln x}{x^2 + 4x + 1} \, dx \]

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

We first rewrite \( I_1 \) using a substitution. Let \( x = \frac{1}{t} \), 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 can rewrite \( I_1 \):

\[ I_1 = \int_\infty^1 \frac{|\ln(\frac{1}{t})|}{(\frac{1}{t})^2 + 4(\frac{1}{t}) + 1} \left(-\frac{1}{t^2}\right) dt \]

### Step 2: Simplifying the integrand

Now, we simplify the integrand:

\[ |\ln(\frac{1}{t})| = -\ln(t) \quad \text{(since \( \ln(\frac{1}{t}) = -\ln(t) \))} \]

The denominator becomes:

\[ \left(\frac{1}{t}\right)^2 + 4\left(\frac{1}{t}\right) + 1 = \frac{1 + 4t + t^2}{t^2} \]

Thus, we have:

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

This simplifies to:

\[ I_1 = \int_1^\infty \frac{\ln(t)}{1 + 4t + t^2} dt \]

### Step 3: Comparing \( I_1 \) and \( I_2 \)

Now we see that:

\[ I_1 = \int_1^\infty \frac{\ln(t)}{1 + 4t + t^2} dt \]

And we have:

\[ I_2 = \int_1^\infty \frac{\ln x}{x^2 + 4x + 1} dx \]

Notice that \( 1 + 4t + t^2 = t^2 + 4t + 1 \). Therefore:

\[ I_1 = I_2 \]

### Conclusion

Thus, we conclude that:

\[ I_1 = I_2 \]