<s>[INST] Evaluating $\int_0^1\frac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx$ 
$$\int_0^1\dfrac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx$$

Attempt:
If we write: $f(x)= x^4 + x^3+ x^2$, we get:
$$I = \displaystyle\int_0^1 \dfrac{3f(x)+x^3}{(f'(x)+1)^2}\, dx$$
I have no idea how to proceed. Integration by parts/ substitution can't help.
I don't need the full solution. Just a guiding hint would suffice.
 [/INST] The hint of @achillehui to substitute $x\to y=\frac{1}{x}$  is valuable and should be an answer by its own.

We obtain
  \begin{align*}
\color{blue}{\int_0^1\frac{3x^4+4x^3+3x^2}{\left(4x^3+3x^2+2x+1\right)^2}\,dx}
&=\int_1^\infty\frac{3y^{-4}+4y^{-3}+3y^{-2}}{\left(4y^{-3}+3y^{-2}+2y^{-1}+1\right)^2}y^{-2}\,dy\tag{1}\\ %fixed denominator here
&=\int_1^\infty\frac{3y^2+4y+3}{\left(y^3+2y^2+3y+4\right)^2}\,dy\tag{2}\\
&=-\left.\frac{1}{y^3+2y^2+3y+4}\right|_1^\infty\tag{3}\\
&\,\,\color{blue}{=\frac{1}{10}}
\end{align*}

Comment:


*

*In (1) we substitute $y=\frac{1}{x},\quad dy=-\frac{1}{x^2}dx$.

*In (2) we expand with $y^6$.

*In (3) we integrate by noting that $\frac{d}{dy}\left(y^3+2y^2+3y+4\right)=3y^2+4y+3$.
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