1. We need to evaluate $I=\int \frac{3x^{13}+2x^{11}}{(2x^4+3x^2+1)^4}\,dx.$

2. First simplify the numerator: $3x^{13}+2x^{11}=x^{11}(3x^2+2).$ So, $I=\int \frac{x^{11}(3x^2+2)}{(2x^4+3x^2+1)^4}\,dx.$

3. Observe the expression in the denominator: $2x^4+3x^2+1.$ Its derivative is $\frac{d}{dx}(2x^4+3x^2+1)=8x^3+6x=2x(4x^2+3),$ which does not directly match the numerator. So we try to rewrite the integral using a substitution involving $x^2$.

Let $t=x^2 \implies dt=2x\,dx.$ Then a more useful substitution is obtained by checking the options.

4. Since the denominator has power $4$, a standard pattern is $\frac{d}{dx}\left(\frac{x^{12}}{(2x^4+3x^2+1)^3}\right)$ or $\frac{d}{dx}\left(\frac{x^4}{(2x^4+3x^2+1)^3}\right).$ Let us test option A directly.

Take $F(x)=\frac{x^{12}}{6(2x^4+3x^2+1)^3}.$ Differentiate using product rule: $F(x)=\frac{1}{6}x^{12}(2x^4+3x^2+1)^{-3}.$ Then $F'(x)=\frac{1}{6}\left[12x^{11}(2x^4+3x^2+1)^{-3}+x^{12}(-3)(2x^4+3x^2+1)^{-4}(8x^3+6x)\right].$

5. Simplify: $F'(x)=2x^{11}(2x^4+3x^2+1)^{-3}-\frac{1}{2}x^{12}(8x^3+6x)(2x^4+3x^2+1)^{-4}.$ Bring to common denominator $(2x^4+3x^2+1)^4$: $F'(x)=\frac{2x^{11}(2x^4+3x^2+1)-\frac{1}{2}x^{12}(8x^3+6x)}{(2x^4+3x^2+1)^4}.$ Now simplify numerator: $2x^{11}(2x^4+3x^2+1)=4x^{15}+6x^{13}+2x^{11},$ and $\frac{1}{2}x^{12}(8x^3+6x)=4x^{15}+3x^{13}.$ Therefore, $\text{numerator}= (4x^{15}+6x^{13}+2x^{11})-(4x^{15}+3x^{13})=3x^{13}+2x^{11}.$ So, $F'(x)=\frac{3x^{13}+2x^{11}}{(2x^4+3x^2+1)^4}.$ This exactly matches the integrand.

6. Hence, $\int \frac{3x^{13}+2x^{11}}{(2x^4+3x^2+1)^4}\,dx=\frac{x^{12}}{6(2x^4+3x^2+1)^3}+C.$

7. Therefore the correct option is: $\boxed{\text{A}}.$

8. Comparison with stored answer: Stored correct answer = A, which matches our derived answer.