Let $I = \int x^{2}\log(x^{2}+1)dx$

We will use integration by parts, $\int u dv = uv - \int v du$.

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

Then $du = \frac{2x}{x^{2}+1}dx$ and $v = \int x^{2}dx = \frac{x^{3}}{3}$.

So, $I = \frac{x^{3}}{3}\log(x^{2}+1) - \int \frac{x^{3}}{3} \cdot \frac{2x}{x^{2}+1}dx$

$I = \frac{x^{3}}{3}\log(x^{2}+1) - \frac{2}{3}\int \frac{x^{4}}{x^{2}+1}dx$

Now, we need to evaluate $\int \frac{x^{4}}{x^{2}+1}dx$. We can rewrite the integrand as:

$\frac{x^{4}}{x^{2}+1} = \frac{x^{4}-1+1}{x^{2}+1} = \frac{(x^{2}-1)(x^{2}+1)+1}{x^{2}+1} = x^{2}-1 + \frac{1}{x^{2}+1}$

So, $\int \frac{x^{4}}{x^{2}+1}dx = \int (x^{2}-1 + \frac{1}{x^{2}+1})dx = \int x^{2}dx - \int dx + \int \frac{1}{x^{2}+1}dx$

$= \frac{x^{3}}{3} - x + \arctan(x) + C_{1}$

Substituting this back into the expression for $I$:

$I = \frac{x^{3}}{3}\log(x^{2}+1) - \frac{2}{3}(\frac{x^{3}}{3} - x + \arctan(x)) + C$

$I = \frac{x^{3}}{3}\log(x^{2}+1) - \frac{2x^{3}}{9} + \frac{2x}{3} - \frac{2}{3}\arctan(x) + C$