Simplify the integrand using polynomial long division or algebraic manipulation.

We have $\frac{x^2}{x+1}$. We can rewrite the numerator to make it divisible by $x+1$.

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

$$\int \left(x-1 + \frac{1}{x+1}\right) dx = \int x\,dx - \int 1\,dx + \int \frac{1}{x+1}\,dx$$ $$= \frac{x^2}{2} - x + \ln|x+1|$$

$$\int_{1}^{3} \frac{x^2}{x+1}\,dx = \left[ \frac{x^2}{2} - x + \ln|x+1| \right]_{1}^{3}$$

Evaluate at the upper limit ($x=3$):

$$\left( \frac{3^2}{2} - 3 + \ln|3+1| \right) = \left( \frac{9}{2} - 3 + \ln(4) \right) = (4.5 - 3 + \ln(4)) = 1.5 + \ln(4)$$

Evaluate at the lower limit ($x=1$):

$$\left( \frac{1^2}{2} - 1 + \ln|1+1| \right) = \left( \frac{1}{2} - 1 + \ln(2) \right) = (0.5 - 1 + \ln(2)) = -0.5 + \ln(2)$$

Subtract the lower limit value from the upper limit value:

$$(1.5 + \ln(4)) - (-0.5 + \ln(2)) = 1.5 + \ln(4) + 0.5 - \ln(2)$$ $$= 2 + \ln(4) - \ln(2)$$

Using the logarithm property $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$:

$$= 2 + \ln\left(\frac{4}{2}\right) = 2 + \ln(2)$$

The final answer is $2 + \ln(2)$.