# Integral $\int_{0}^{1}\frac{3x+4}{x^3-3x-4}\mathrm{d}x$

Evaluate the integral $$\int_{0}^{1}\frac{3x+4}{x^3-3x-4}\mathrm{d}x$$

What should be the approach for evaluating this integral? No substitution seems to be working and the denominator also doesn't have good roots so partial fraction method would be too tedious.

• @Surb What substitution? – Light Yagami Jul 16 '20 at 10:42
• Sorry : Only fractional decomposition works here. And the result is anyway not nice – Surb Jul 16 '20 at 10:43
• Try substituting $x \rightarrow y+1/y$, it'll make the denominator easier to factorise. – Nikunj Jul 16 '20 at 10:47
• Okay, but how to do fractional decomposition, can you show that? I found out the roots of the cubic to be $2\cos(\alpha),2\cos(\frac{2\pi}{3}\pm \alpha)$, where $\alpha=\frac{\cos^{-1}(2)}{3}$ – Light Yagami Jul 16 '20 at 10:47
• @Clayton I wrote it correct. The roots are complex. – Light Yagami Jul 16 '20 at 11:14

## 2 Answers

The computation is manageable.

The denominator can be factored as

$$x^3-3x-4=(x-a)(x^2+ax+c)$$

where $$a$$ is the real root, and

$$c=a^2-3=\dfrac4a.$$

Then if you decompose the function as

$$\frac{2A}{x-a}-\frac{A(2x+a)}{x^2+ax+c}+\frac B{x^2+ax+c},$$

every term is "easy" to integrate. By identification,

$$\begin{cases}3Aa+B=-3,\\3A(a^2-2)-aB=-4.\end{cases}$$

• a little typo , it's $x^3$ instead of $x^2$ in the first line. – Light Yagami Jul 16 '20 at 11:20
• @ABCD: yep, fixed. – Yves Daoust Jul 16 '20 at 11:22

Here is a possible way to proceed. The roots of the denominator inside the integral are... \begin{aligned} a_1 &= \sqrt{2+\sqrt 3} + \sqrt{2-\sqrt 3}\ ,\\ a_2 &= u\sqrt{2+\sqrt 3} + u^2\sqrt{2-\sqrt 3}\ ,\\ a_3 &= u^2\sqrt{2+\sqrt 3} + u\sqrt{2-\sqrt 3}\ . \\[2mm] &\qquad\text{Here u is a primitive third root of the unit,}\\ u &=\frac 12(-1+i\sqrt 3) \end{aligned} Indeed, all of them are of the shape $$(b+c)$$ where $$b^3, c^3$$ are $$(2\pm \sqrt 3)$$, and we compute: $$(b+c)^3=(b^3+c^3)+3bc(b+c)=(2+2)+3\cdot1\cdot(b+c)\ .$$ (So $$(b+c)$$ is a root of the equation $$x^3=3x+4$$ in all three cases.)

We will write in the sequel expressions, which are symmetric w.r.t. permutations of the set $$A=\{a_1,a_2,a_3\}$$, and when writing sums over $$A$$ we use $$a$$ as running index. Observe that for some polynomial $$p$$ of degree $$<3$$ we have $$\frac{p(x)}{x^3-3x-4} =\sum_{a\in A}\frac{p(a)}{3a^2-3}\cdot \frac 1{x-a}\ .$$ Indeed, we multiply with $$(x^3-3x-4)=\prod(x-a)$$, and check the equality of polynomials of degree $$<3$$ in the three places $$a\in A$$. This reduces to the computation of the limit of $$(x^3-3x-4)/(x-a)$$ for $$x\to a$$, so we take the derivative of $$(x^3-3x-4)$$ in $$a$$, which is $$(3a^2-3)$$.

In our case: $$\frac{3x+4}{x^3-3x-4} =\frac 13\sum_{a\in A}\frac{3a+4}{a^2-1}\cdot\frac 1{x-a}\ .$$ We integrate now from $$0$$ to $$1$$, getting $$\frac 13\sum_{a\in A}\frac{3a+4}{a^2-1}\cdot\log\frac{1-a}{0-a} = \frac 1{18}\sum_{a\in A}(2a^2+5a-4)\cdot\log\frac{1-a}{0-a}\ .$$

Here are some computer lines written in sage, supporting the above.

sage: a = (2+sqrt(3))^(1/3) + (2-sqrt(3))^(1/3)
sage: a.minpoly()
x^3 - 3*x - 4
sage: a.n()
2.19582334544565

sage: roots = (x^3-3*x-4).roots(ring=QQbar, multiplicities=False)
sage: roots    [2.195823345445648?,
-1.097911672722824? - 0.7850032632435902?*I,
-1.097911672722824? + 0.7850032632435902?*I]
sage: # note that the above numbers are "exact" and live in QQbar

sage: R.<x> = PolynomialRing(QQbar)
sage: sum([ (3*a+4)/3/(a^2-1) / (x-a) for a in roots ]) == (3*x+4)/(x^3-3*x-4)
True
sage: integral((3*x+4)/(x^3-3*x-4), x, 0, 1).n()
-1.04297088692137
sage: sum( [ (3*a+4)/3/(a^2-1) * (log(1-a) - log(0-a)) for a in roots ] ).n()
-1.04297088748547

sage: R.<x> = PolynomialRing(QQ)
sage: K.<a> = NumberField(x^3 - 3*x - 4, embedding=2.19)
sage: K
Number Field in a with defining polynomial x^3 - 3*x - 4 with a = 2.195823345445648?
sage: (3*a+4) / (3*a^2-3)
1/9*a^2 + 5/18*a - 2/9

sage: sum( [ (2*a^2+5*a-4)/18 * (log(1-a) - log(0-a)) for a in roots ] ).n()
-1.04297088748547