$$\int_0^1\dfrac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx$$
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.
On
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 $\frac{d}{dy}\left(y^3+2y^2+3y+4\right)=3y^2+4y+3$.
Since this integral is very closely linked to the Quotient Rule for differentiation (due to the square in the denominator), we will try to write the numerator as an expression of the form $$P'(x)\cdot(4x^3+3x^2+2x+1)-P(x)\cdot(4x^3+3x^2+2x+1)'$$ so $$3x^4+4x^3+3x^2=P'(x)\cdot(4x^3+3x^2+2x+1)-2P(x)\cdot(6x^2+3x+1).$$
We see that $P$ must be at least a quadratic, so $P(x)=ax^2+bx+c$ for some real numbers $a,b,c$. Then $$\begin{align}3x^4+4x^3+3x^2&=(2ax+b)\cdot(4x^3+3x^2+2x+1)-2(ax^2+bx+c)(6x^2+3x+1)\\&=8ax^4+(6a+4b)x^3+(4a+3b)x^2+(2a+b)x+b-12ax^4-(6a+12b)x^3\\&\,\,\,\,\,\,-(2a+6b+12c)x^2-(2b+6c)x-2c\\&=-4ax^4-8bx^3+(2a-3b-9c)x^2+(2a-b-6c)x+b-2c\end{align}$$ Hence $$a=-\frac34,\quad b=-\frac12,\quad c=-\frac14$$ This means that $$P(x)=-\frac14(3x^2+2x+1)$$ so $$\int_0^1\dfrac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx=-\frac14\left[\frac{3x^2+2x+1}{4x^3 + 3x^2 + 2x+ 1}\right]_0^1=\frac1{10}.$$