The integral $$\int \frac{\sin^2 x \cos^2 x}{(\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x)^2} dx$$ is equal to
(where C is the constant of integration).

We begin with the integral

$$I=\int \dfrac{\sin^2 x\,\cos^2 x}{(\sin^5 x+\cos^3 x\sin^2 x+\sin^3 x\cos^2 x+\cos^5 x)^2}\,dx.$$

First we simplify the rather complicated expression that appears in the denominator. We write every term in powers of $$\sin x$$ and $$\cos x$$ and then factor out the highest common power of $$\cos x$$:

$$\sin^5 x+\cos^3 x\sin^2 x+\sin^3 x\cos^2 x+\cos^5 x =\cos^5 x\Bigl(\tan^5 x+\tan^2 x+\tan^3 x+1\Bigr).$$

Inside the large parentheses we recognise a product. Indeed,

$$\bigl(1+\tan^3 x\bigr)\bigl(1+\tan^2 x\bigr)=1+\tan^2 x+\tan^3 x+\tan^5 x,$$

which is exactly the bracket above. Hence we may write

$$\sin^5 x+\cos^3 x\sin^2 x+\sin^3 x\cos^2 x+\cos^5 x =\cos^5 x\,(1+\tan^3 x)\,(1+\tan^2 x).$$

Because $$1+\tan^2 x=\sec^2 x,$$ the entire factor becomes

$$\cos^5 x\,(1+\tan^3 x)\,\sec^2 x =\cos^3 x\,(1+\tan^3 x),$$

and therefore the square that appears in the denominator of the integrand is

$$\bigl(\sin^5 x+\cos^3 x\sin^2 x+\sin^3 x\cos^2 x+\cos^5 x\bigr)^2 =\cos^6 x\,(1+\tan^3 x)^2.$$

Now we replace every occurrence of $$\sin x$$ and $$\cos x$$ in the integrand:

$$\dfrac{\sin^2 x\;\cos^2 x}{\cos^6 x\,(1+\tan^3 x)^2} =\dfrac{\sin^2 x\;\cos^2 x}{\cos^6 x}\, \dfrac{1}{(1+\tan^3 x)^2} =\dfrac{\tan^2 x\cos^4 x}{\cos^6 x}\, \dfrac{1}{(1+\tan^3 x)^2} =\dfrac{\tan^2 x}{\cos^2 x}\, \dfrac{1}{(1+\tan^3 x)^2}.$$

Because $$\dfrac{1}{\cos^2 x}=\sec^2 x=1+\tan^2 x,$$ the integrand becomes

$$\dfrac{\tan^2 x\bigl(1+\tan^2 x\bigr)}{(1+\tan^3 x)^2}.$$

At this stage it is very natural to introduce the substitution

$$t=\tan x.$$

We know the basic differentiation formula $$\dfrac{d}{dx}(\tan x)=\sec^2 x=1+t^2,$$ so

$$dt=(1+t^2)\,dx\quad\Longrightarrow\quad dx=\dfrac{dt}{1+t^2}.$$

Substituting for both the integrand and $$dx$$ we obtain

$$I=\int \dfrac{t^2\,(1+t^2)}{(1+t^3)^2}\;\dfrac{dt}{1+t^2} =\int \dfrac{t^2}{(1+t^3)^2}\,dt.$$

The fraction has now become a very clean rational function in $$t$$. We recognise the numerator $$t^2\,dt$$ as essentially the differential of $$t^3$$. Indeed,

$$\dfrac{d}{dt}\bigl(t^3\bigr)=3t^2.$$

So we set

$$u=1+t^3\quad\Longrightarrow\quad du=3t^2\,dt \quad\Longrightarrow\quad t^2\,dt=\dfrac{du}{3}.$$

Substituting $$u$$ and $$du$$ into the integral we get

$$I=\int \dfrac{1}{3}\,\dfrac{du}{u^2} =\dfrac{1}{3}\int u^{-2}\,du.$$

We recall the power-rule formula $$\displaystyle\int u^n\,du=\dfrac{u^{\,n+1}}{n+1}+C$$ for every exponent $$n\neq-1$$. Here $$n=-2$$, so

$$\int u^{-2}\,du=\dfrac{u^{-1}}{-1}+C=-u^{-1}+C.$$

Putting the numeric factor back,

$$I=\dfrac{1}{3}\bigl(-u^{-1}\bigr)+C=-\dfrac{1}{3u}+C.$$

Finally we reverse all substitutions. First $$u=1+t^3$$ and then $$t=\tan x$$, giving

$$I=-\dfrac{1}{3\bigl(1+\tan^3 x\bigr)}+C.$$

This matches Option C in the list provided.

Hence, the correct answer is Option C.