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$$\int_0^1\dfrac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx$$

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.

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    $\begingroup$ Hint: change variable from $x \to y = \frac1x$. In the new variable, when you express the integrand as $\frac{p(y)}{q(y)^2}$, what is the relation between $p(y)$ and $q(y)$? Once you do that, the integral trivally evaluates to $\frac{1}{10}$. $\endgroup$
    – achille hui
    Jul 5, 2018 at 8:29
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    $\begingroup$ @samjoe just like last time, OP will figure that out and probably post an answer.... $\endgroup$
    – achille hui
    Jul 5, 2018 at 8:50
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    $\begingroup$ @achille I respect your opinion, but my say on this is that your comment is less of a comment but a full answer. So rather it made more sense to post it as such. Same with last question which I just saw. But is is obviously your call not mine. $\endgroup$
    – jonsno
    Jul 5, 2018 at 9:50
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    $\begingroup$ If all other “shortcut” methods don’t work, there is always the brute force method that involves factoring and then partial fraction expansion. $\endgroup$
    – Colin Hicks
    Jul 5, 2018 at 12:35

2 Answers 2

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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 $$\int_0^1\dfrac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx=-\frac14\left[\frac{3x^2+2x+1}{4x^3 + 3x^2 + 2x+ 1}\right]_0^1=\frac1{10}.$$

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    $\begingroup$ Amazing! Thanks. $\endgroup$
    – Archer
    Jul 5, 2018 at 12:38
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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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