To solve the integral of $\int \frac{\ln(1 + x^2)}{1 + x^2} dx$ using contour integration, let's proceed with a detailed breakdown of the approach.

We are asked to use the contour integral:

$J = \oint \frac{\ln(z) + i}{z^2 + 1} \, dz$

where the contour is a path from $-R$ to $+R$ along the real axis, and then a semicircular arc from $R$ back to $-R$ above the real axis. This contour will help us solve the given integral using complex analysis. Here’s the step-by-step process.

Step 1: Identifying the Integrand and Contour

The function in the contour integral is:

$f(z) = \frac{\ln(z) + i}{z^2 + 1}$

The poles of the function are the points where the denominator is zero. Since $z^2 + 1 = 0$, the poles are located at $z = i$ and $z = -i$.

Step 2: Residue Calculation

To apply the residue theorem, we need to compute the residues of $f(z)$ inside the contour. Since the contour goes above the real axis, it will enclose the pole at $z = i$ but not the pole at $z = -i$.

Let’s calculate the residue of $f(z)$ at $z = i$.

$\text{Residue of } f(z) \text{ at } z = i$

For this, we first rewrite $f(z)$ in terms of partial fractions near $z = i$. The denominator $z^2 + 1 = (z - i)(z + i)$, so we can express the residue as:

$\text{Res}(f(z), z = i) = \lim_{z \to i} (z - i) \frac{\ln(z) + i}{(z - i)(z + i)} = \frac{\ln(i) + i}{2i}$

Using $\ln(i) = i\frac{\pi}{2}$, we find:

$\text{Res}(f(z), z = i) = \frac{i\frac{\pi}{2} + i}{2i} = \frac{\pi}{4} + \frac{1}{2}$

Step 3: Applying the Residue Theorem

The residue theorem states that for a contour integral around a closed loop, the integral is $2\pi i$ times the sum of the residues inside the contour. Therefore, we have:

$J = 2\pi i \left( \frac{\pi}{4} + \frac{1}{2} \right)$

Step 4: Real Axis Integral

Now, split the contour integral into two parts: the real axis part and the semicircular arc part. The contribution from the semicircular arc goes to zero as $R \to \infty$, leaving:

$\int_{-\infty}^{\infty} \frac{\ln(1 + x^2)}{1 + x^2} \, dx = \pi \left( \frac{\pi}{4} + \frac{1}{2} \right)$

Simplifying the result:

$\int_{-\infty}^{\infty} \frac{\ln(1 + x^2)}{1 + x^2} \, dx = \frac{\pi^2}{4} + \frac{\pi}{2}$

Final Answer

Thus, the value of the integral is:

$\int_{-\infty}^{\infty} \frac{\ln(1 + x^2)}{1 + x^2} \, dx = \frac{\pi^2}{4} + \frac{\pi}{2}$

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Would you like to explore further details on contour integration or related topics? Here are some questions you may find interesting:

1. How do you handle integrals with branch cuts in contour integration?
2. What happens if the contour encloses both poles at $z = i$ and $z = -i$?
3. Can the same method be applied to other integrals involving logarithms or rational functions?
4. How does the semicircular arc's contribution vanish as $R \to \infty$?
5. What are the different ways to calculate residues in complex analysis?

Tip: In contour integration, always carefully consider the nature of the singularities and use the residue theorem efficiently to evaluate integrals along the real axis.