The integral $$\int \frac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4} dx$$, is equal to

We have to evaluate the integral

$$\int \dfrac{3x^{13}+2x^{11}}{(2x^4+3x^2+1)^4}\,dx.$$

By inspection, the denominator is a power of the polynomial $$2x^4+3x^2+1.$$ A very common technique is to look for a function whose derivative would give exactly a numerator of the form “(polynomial in $$x$$) divided by $$(2x^4+3x^2+1)^4$$.” Experience suggests differentiating a candidate of the type $$\dfrac{x^{m}}{(2x^4+3x^2+1)^3},$$ because lowering the power of the denominator by one during differentiation normally produces the power four that we see in the integrand.

So, let us consider the function

$$F(x)=\dfrac{x^{12}}{(2x^4+3x^2+1)^3}.$$

Before differentiating, we explicitly state the formulas we shall use:

1. Power rule: $$\dfrac{d}{dx}\,x^{n}=n\,x^{\,n-1}.$$

2. Quotient / product rule in disguise: writing $$F(x)=x^{12}\bigl(2x^4+3x^2+1\bigr)^{-3}$$, we apply the product rule $$\dfrac{d}{dx}\bigl(u\,v\bigr)=u'\,v+u\,v'.$$

Differentiating step by step, we get

$$\begin{aligned} F'(x)&=\dfrac{d}{dx}\Bigl[x^{12}\bigl(2x^4+3x^2+1\bigr)^{-3}\Bigr] \\ &=\left(\dfrac{d}{dx}x^{12}\right)\bigl(2x^4+3x^2+1\bigr)^{-3} +x^{12}\,\dfrac{d}{dx}\Bigl[\bigl(2x^4+3x^2+1\bigr)^{-3}\Bigr]. \end{aligned}$$

Using the power rule on $$x^{12}$$ we have $$12x^{11}$$. For the second derivative we recall the chain rule: if $$h(x)=(g(x))^{-3}$$ then $$h'(x)=-3(g(x))^{-4}g'(x)$$. Here $$g(x)=2x^4+3x^2+1$$, and

$$g'(x)=\dfrac{d}{dx}(2x^4)+\dfrac{d}{dx}(3x^2)+\dfrac{d}{dx}(1)=8x^3+6x.$$

Applying this,

$$\begin{aligned} F'(x) &=12x^{11}\,(2x^4+3x^2+1)^{-3} +x^{12}\,\bigl[-3(2x^4+3x^2+1)^{-4}\,(8x^3+6x)\bigr] \\ &=\dfrac{12x^{11}}{(2x^4+3x^2+1)^{3}} -\dfrac{3x^{12}(8x^3+6x)}{(2x^4+3x^2+1)^{4}}. \end{aligned}$$

To combine these two fractions we convert the first term so that it has the same fourth-power denominator:

$$\dfrac{12x^{11}}{(2x^4+3x^2+1)^{3}} =\dfrac{12x^{11}\bigl(2x^4+3x^2+1\bigr)}{(2x^4+3x^2+1)^{4}} =\dfrac{12x^{11}(2x^4+3x^2+1)}{(2x^4+3x^2+1)^{4}}.$$

Now expand the numerator:

$$12x^{11}(2x^4+3x^2+1)=24x^{15}+36x^{13}+12x^{11}.$$

The second term’s numerator is

$$-3x^{12}(8x^3+6x)=-(24x^{15}+18x^{13}).$$

Adding the numerators:

$$\begin{aligned} (24x^{15}+36x^{13}+12x^{11})+(-24x^{15}-18x^{13}) &=(24x^{15}-24x^{15})+(36x^{13}-18x^{13})+12x^{11} \\ &=18x^{13}+12x^{11}. \end{aligned}$$

Hence

$$F'(x)=\dfrac{18x^{13}+12x^{11}}{(2x^4+3x^2+1)^{4}} =\dfrac{6(3x^{13}+2x^{11})}{(2x^4+3x^2+1)^{4}}.$$

Dividing both sides by $$6$$ gives

$$\dfrac{3x^{13}+2x^{11}}{(2x^4+3x^2+1)^{4}} =\dfrac{1}{6}\,F'(x).$$

Therefore the integrand is exactly $$\dfrac{1}{6}$$ times the derivative of $$F(x)$$. So, integrating is immediate:

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

Writing the result cleanly,

$$\boxed{\dfrac{x^{12}}{6\,(2x^4+3x^2+1)^{3}}+C}.$$

Hence, the correct answer is Option D.