The problem asks us to evaluate the definite integral $\int_{0}^{1} \log(x^2 + 1) \, dx$.

1. Problem Description

The problem asks us to evaluate the definite integral $\int_{0}^{1} \log(x^2 + 1) \, dx$.

2. Solution Steps

We can solve this integral using integration by parts.

Recall that integration by parts is given by the formula:

$\int u \, dv = uv - \int v \, du$

Let $u = \log(x^2 + 1)$ and $dv = dx$.

Then $du = \frac{2x}{x^2 + 1} dx$ and $v = x$.

Therefore,

$\int_{0}^{1} \log(x^2 + 1) \, dx = \left[x \log(x^2 + 1)\right]_{0}^{1} - \int_{0}^{1} x \cdot \frac{2x}{x^2 + 1} \, dx$

$= \left[1 \cdot \log(1^2 + 1) - 0 \cdot \log(0^2 + 1)\right] - \int_{0}^{1} \frac{2x^2}{x^2 + 1} \, dx$

$= \log(2) - \int_{0}^{1} \frac{2x^2}{x^2 + 1} \, dx$

Now we need to evaluate $\int_{0}^{1} \frac{2x^2}{x^2 + 1} \, dx$.

We can rewrite the integrand as follows:

$\frac{2x^2}{x^2 + 1} = \frac{2(x^2 + 1) - 2}{x^2 + 1} = 2 - \frac{2}{x^2 + 1}$

Thus, $\int_{0}^{1} \frac{2x^2}{x^2 + 1} \, dx = \int_{0}^{1} \left(2 - \frac{2}{x^2 + 1}\right) \, dx$

$= \int_{0}^{1} 2 \, dx - \int_{0}^{1} \frac{2}{x^2 + 1} \, dx$

$= \left[2x\right]_{0}^{1} - 2 \left[\arctan(x)\right]_{0}^{1}$

$= (2(1) - 2(0)) - 2 (\arctan(1) - \arctan(0))$

$= 2 - 2 \left(\frac{\pi}{4} - 0\right) = 2 - \frac{\pi}{2}$

Substituting this back into our original expression:

$\int_{0}^{1} \log(x^2 + 1) \, dx = \log(2) - \left(2 - \frac{\pi}{2}\right)$

$= \log(2) - 2 + \frac{\pi}{2}$

$= \log(2) + \frac{\pi}{2} - 2$

3. Final Answer

$\log(2) + \frac{\pi}{2} - 2$