$$\lim_{n \to \infty} \left(\frac{n^2}{(n^2+1)(n+1)} + \frac{n^2}{(n^2+4)(n+2)} + \frac{n^2}{(n^2+9)(n+3)} + \cdots + \frac{n^2}{(n^2+n^2)(n+n)}\right)$$ is equal to

We need to evaluate:

$$\lim_{n \to \infty} \sum_{r=1}^{n} \frac{n^2}{(n^2+r^2)(n+r)}$$

We rewrite the general term by factoring out $$n^3$$ from the denominator:

$$\frac{n^2}{(n^2+r^2)(n+r)} = \frac{n^2}{n^2\left(1+\frac{r^2}{n^2}\right) \cdot n\left(1+\frac{r}{n}\right)} = \frac{1}{n} \cdot \frac{1}{\left(1+\left(\frac{r}{n}\right)^2\right)\left(1+\frac{r}{n}\right)}$$

Let $$x = \frac{r}{n}$$. As $$n \to \infty$$, the Riemann sum becomes:

$$I = \int_0^1 \frac{dx}{(1+x^2)(1+x)}$$

We perform partial fraction decomposition. We write:

$$\frac{1}{(1+x)(1+x^2)} = \frac{A}{1+x} + \frac{Bx+C}{1+x^2}$$

Multiplying both sides by $$(1+x)(1+x^2)$$:

$$1 = A(1+x^2) + (Bx+C)(1+x)$$

Substituting $$x = -1$$: $$1 = A(1+1) + 0 = 2A$$, so $$A = \frac{1}{2}$$.

Expanding the right side: $$1 = A + Ax^2 + Bx + Bx^2 + C + Cx$$

$$= (A+C) + (B+C)x + (A+B)x^2$$

Comparing coefficients of $$x^2$$: $$0 = A + B = \frac{1}{2} + B$$, so $$B = -\frac{1}{2}$$.

Comparing constant terms: $$1 = A + C = \frac{1}{2} + C$$, so $$C = \frac{1}{2}$$.

Therefore:

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

We can verify: $$\frac{1}{2} \cdot \frac{1-x}{1+x^2} = \frac{1}{2} \cdot \frac{1}{1+x^2} - \frac{1}{2} \cdot \frac{x}{1+x^2}$$

Now integrating each term from 0 to 1:

First integral: $$\frac{1}{2}\int_0^1 \frac{dx}{1+x} = \frac{1}{2}[\ln(1+x)]_0^1 = \frac{1}{2}(\ln 2 - \ln 1) = \frac{1}{2}\ln 2$$

Second integral: $$\frac{1}{2}\int_0^1 \frac{dx}{1+x^2} = \frac{1}{2}[\tan^{-1}x]_0^1 = \frac{1}{2}\left(\frac{\pi}{4} - 0\right) = \frac{\pi}{8}$$

Third integral: $$-\frac{1}{2}\int_0^1 \frac{x\,dx}{1+x^2}$$

Let $$u = 1+x^2$$, then $$du = 2x\,dx$$:

$$= -\frac{1}{2} \cdot \frac{1}{2}[\ln(1+x^2)]_0^1 = -\frac{1}{4}(\ln 2 - \ln 1) = -\frac{1}{4}\ln 2$$

Adding all three parts:

$$I = \frac{1}{2}\ln 2 + \frac{\pi}{8} - \frac{1}{4}\ln 2 = \frac{\pi}{8} + \frac{1}{4}\ln 2$$

The correct answer is Option A.