1. Rewrite the summand in a Riemann-sum form

We need to evaluate

$$L=\lim_{n\to\infty}\sum_{k=1}^n \frac{n^3}{(n^2+k^2)(n^2+3k^2)}.$$

Factor out $n^2$ from each bracket:

$$(n^2+k^2)=n^2\left(1+\left(\frac{k}{n}\right)^2\right),$$ $$(n^2+3k^2)=n^2\left(1+3\left(\frac{k}{n}\right)^2\right).$$

So the summand becomes

$$\frac{n^3}{n^4\left(1+(k/n)^2\right)\left(1+3(k/n)^2\right)} =\frac{1}{n}\cdot \frac{1}{\left(1+(k/n)^2\right)\left(1+3(k/n)^2\right)}.$$

Hence

$$L=\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{n} \frac{1}{\left(1+(k/n)^2\right)\left(1+3(k/n)^2\right)}.$$

This is the Riemann sum for

$$L=\int_0^1 \frac{dx}{(1+x^2)(1+3x^2)}.$$

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2. Use partial fractions

Let

$$\frac{1}{(1+x^2)(1+3x^2)}=\frac{A}{1+x^2}+\frac{B}{1+3x^2}.$$

Then

$$1=A(1+3x^2)+B(1+x^2).$$

So,

$$1=(A+B)+(3A+B)x^2.$$

Comparing coefficients:

$$A+B=1,$$ $$3A+B=0.$$

Subtracting,

$$2A=-1\Rightarrow A=-\frac12,$$

and therefore

$$B=\frac32.$$

Thus,

$$\frac{1}{(1+x^2)(1+3x^2)}=-\frac{1}{2(1+x^2)}+\frac{3}{2(1+3x^2)}.$$

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3. Integrate term by term

Therefore,

$$L=\int_0^1\left(-\frac{1}{2(1+x^2)}+\frac{3}{2(1+3x^2)}\right)dx.$$

Now,

$$\int \frac{dx}{1+x^2}=\tan^{-1}x,$$

and

$$\int \frac{dx}{1+3x^2}=\frac{1}{\sqrt3}\tan^{-1}(\sqrt3 x).$$

So,

$$L=-\frac12\left[\tan^{-1}x\right]_0^1+ \frac32\cdot \frac{1}{\sqrt3}\left[\tan^{-1}(\sqrt3 x)\right]_0^1.$$

Evaluate limits:

$$\tan^{-1}(1)=\frac\pi4, \qquad \tan^{-1}(\sqrt3)=\frac\pi3.$$

Hence

$$L=-\frac12\cdot \frac\pi4+\frac{3}{2\sqrt3}\cdot \frac\pi3.$$ $$L=-\frac\pi8+\frac\pi{2\sqrt3}.$$

Now simplify:

$$L=\pi\left(\frac{1}{2\sqrt3}-\frac18\right) =\pi\left(\frac{4-\sqrt3}{8\sqrt3}\right).$$

Rationalizing,

$$L=\pi\cdot \frac{\sqrt3(4-\sqrt3)}{24} =\pi\cdot \frac{4\sqrt3-3}{24}.$$

So,

$$L=\frac{(4\sqrt3-3)\pi}{24}.$$

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4. Match with the options

Check option C:

$$\frac{13\pi}{8(4\sqrt3+3)}.$$

Rationalize:

$$\frac{13\pi}{8(4\sqrt3+3)}\cdot \frac{4\sqrt3-3}{4\sqrt3-3} =\frac{13(4\sqrt3-3)\pi}{8\left((4\sqrt3)^2-3^2\right)} =\frac{13(4\sqrt3-3)\pi}{8(48-9)} =\frac{13(4\sqrt3-3)\pi}{8\cdot 39}.$$

Since $13/39=1/3$,

$$=\frac{(4\sqrt3-3)\pi}{24}.$$

This is exactly our value.

Therefore, the correct option is C.

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5. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.