The integral $$\int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3} dx$$, is equal to

We begin with the integral

$$I=\int \dfrac{2x^{12}+5x^{9}}{(x^{5}+x^{3}+1)^{3}}\;dx.$$

The denominator suggests that the primitive should contain the factor $$(x^{5}+x^{3}+1)^{-2},$$ because when we differentiate such a term the exponent will drop from $$-2$$ to $$-3,$$ matching the power $$-3$$ already present in the integrand. Therefore we look for a function of the type

$$F(x)=k\,x^{m}(x^{5}+x^{3}+1)^{-2},$$

where $$k$$ and $$m$$ are constants to be determined. Our aim is to choose $$k$$ and $$m$$ so that $$F'(x)$$ reproduces the numerator $$2x^{12}+5x^{9}.$$

Let us pick the specific trial function

$$F(x)=\dfrac{x^{10}}{2}\,(x^{5}+x^{3}+1)^{-2} \;=\;\dfrac{x^{10}}{2\,(x^{5}+x^{3}+1)^{2}}.$$

Now we differentiate $$F(x)$$. Throughout the calculation we use the following standard formulas:

1. The product rule: $$\dfrac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).$$

2. The chain rule for negative powers: $$\dfrac{d}{dx}\bigl[(g(x))^{-2}\bigr] = -2\,(g(x))^{-3}\,g'(x).$$

Write $$F(x)$$ as a product:

$$F(x)=\dfrac{1}{2}\,x^{10}\,(x^{5}+x^{3}+1)^{-2}.$$

Here $$u(x)=\dfrac{1}{2}x^{10},\quad v(x)=(x^{5}+x^{3}+1)^{-2}.$$

Compute each derivative separately:

$$u'(x)=\dfrac{d}{dx}\Bigl(\dfrac{1}{2}x^{10}\Bigr)=\dfrac{1}{2}\cdot10x^{9}=5x^{9}.$$

For $$v(x),$$ let $$g(x)=x^{5}+x^{3}+1,$$ so that $$v(x)=g(x)^{-2}.$$

Then $$g'(x)=\dfrac{d}{dx}(x^{5}+x^{3}+1)=5x^{4}+3x^{2},$$

and by the chain rule

$$v'(x)=-2\$$, $$g(x)^{-3}\$$, $$g'(x)=-2\$$, $$(x^{5}+x^{3}+1)^{-3}\$$, $$(5x^{4}+3x^{2}).$$

Now apply the product rule:

$$\begin{aligned} F'(x)&=u'(x)v(x)+u(x)v'(x)\\[4pt] &=5x^{9}(x^{5}+x^{3}+1)^{-2}+\frac{x^{10}}{2}\Bigl[-2\,(x^{5}+x^{3}+1)^{-3}(5x^{4}+3x^{2})\Bigr]. \end{aligned}$$

Simplify term by term.

The first term is already simple:

$$5x^{9}(x^{5}+x^{3}+1)^{-2}.$$

In the second term, the factors $$\frac{x^{10}}{2}$$ and $$-2$$ cancel:

$$\frac{x^{10}}{2}\cdot(-2)= -x^{10},$$

so the second term becomes

$$-x^{10}(x^{5}+x^{3}+1)^{-3}(5x^{4}+3x^{2}).$$

Combine the two pieces under the common power $$-3$$. Multiply the first term by an extra factor of $$(x^{5}+x^{3}+1)/(x^{5}+x^{3}+1)$$ to equalize the powers:

$$\begin{aligned} F'(x)&=\Bigl[5x^{9}(x^{5}+x^{3}+1)\Bigr](x^{5}+x^{3}+1)^{-3}-\Bigl[x^{10}(5x^{4}+3x^{2})\Bigr](x^{5}+x^{3}+1)^{-3}\\[4pt] &=\dfrac{5x^{9}(x^{5}+x^{3}+1)-x^{10}(5x^{4}+3x^{2})}{(x^{5}+x^{3}+1)^{3}}. \end{aligned}$$

Now expand the numerators completely:

$$5x^{9}(x^{5}+x^{3}+1)=5x^{14}+5x^{12}+5x^{9},$$

$$x^{10}(5x^{4}+3x^{2})=5x^{14}+3x^{12}.$$

Subtract the two expressions:

$$\bigl[5x^{14}+5x^{12}+5x^{9}\bigr]-\bigl[5x^{14}+3x^{12}\bigr]=2x^{12}+5x^{9}.$$

Thus

$$F'(x)=\dfrac{2x^{12}+5x^{9}}{(x^{5}+x^{3}+1)^{3}}.$$

But the right-hand side is precisely the integrand $$(2x^{12}+5x^{9})/(x^{5}+x^{3}+1)^{3}.$$ Therefore

$$\int \dfrac{2x^{12}+5x^{9}}{(x^{5}+x^{3}+1)^{3}}\;dx = F(x)+C =\dfrac{x^{10}}{2\,(x^{5}+x^{3}+1)^{2}}+C.$$

This matches option D exactly. Hence, the correct answer is Option D.