Evaluate the integral: I = integral dx/((x³+x+1)³)
Evaluate the integral: $$ I = \int \frac{dx}{\left(x^3+x+1\right)^3}. $$
Explanation of Method
Unlike every other integral in this collection, the cubic $x^3+x+1$ does not factor over the rationals — its real root is an irrational algebraic number with no simpler closed form than Cardano's formula. The integral is still, in principle, elementary (every rational function has an elementary antiderivative), but writing out the explicit partial-fraction constants in terms of that irrational root is impractical to carry out by hand. This page documents the correct structural method rather than a fully expanded closed form.
- Confirm $x^3+x+1$ is irreducible over the rationals.
- Identify its unique real root $\alpha$ via Cardano's formula.
- Factor the cubic exactly as $(x-\alpha)(x^2+\alpha x+\alpha^2+1)$.
- Describe the resulting partial-fraction structure.
Step 1 · Irreducibility
By the Rational Root Theorem, the only candidate rational roots of $x^3+x+1$ are $\pm1$, and neither satisfies the equation ($1^3+1+1=3$, $(-1)^3+(-1)+1=-1$). So $x^3+x+1$ has no rational root and is irreducible over $\mathbb Q$. It has exactly one real root $\alpha$ and a complex-conjugate pair, since its discriminant $-4(1)^3-27(1)^2=-31<0$.
Step 2 · The exact real root
By Cardano's formula for $x^3+px+q=0$ with $p=1,\ q=1$:
$$ \alpha = \sqrt[3]{-\frac12+\sqrt{\frac{31}{108}}} + \sqrt[3]{-\frac12-\sqrt{\frac{31}{108}}} \approx -0.68233. $$
Step 3 · Exact factorization
Since $\alpha^3=-\alpha-1$, polynomial division gives exactly:
$$ x^3+x+1 = (x-\alpha)\left(x^2+\alpha x+\alpha^2+1\right), $$
and the quadratic factor is irreducible for every real $\alpha$, since its discriminant $\alpha^2-4(\alpha^2+1)=-3\alpha^2-4$ is always negative.
Step 4 · Partial fraction structure
$$ \frac{1}{(x^3+x+1)^3} = \frac{1}{(x-\alpha)^3\left(x^2+\alpha x+\alpha^2+1\right)^3} = \sum_{k=1}^3\frac{A_k}{(x-\alpha)^k} + \sum_{k=1}^3\frac{B_kx+C_k}{\left(x^2+\alpha x+\alpha^2+1\right)^k}. $$
Clearing denominators produces a single polynomial identity of degree 8 in $x$ with 9 unknown constants $A_1,A_2,A_3,B_1,C_1,B_2,C_2,B_3,C_3$, all of which must be expressed in terms of $\alpha$ (using $\alpha^3=-\alpha-1$ repeatedly to reduce every power of $\alpha$ above degree 2). Solving this system and then integrating each of the six resulting terms — three via the power rule / logarithm for the $(x-\alpha)^k$ pieces, three via the reduction formula for $\int(Bx+C)/(x^2+\alpha x+\alpha^2+1)^k\,dx$ — is a lengthy but entirely mechanical computation, most practically carried out with a computer algebra system rather than by hand.
Summary Table
| Quantity | Role |
|---|---|
| $\alpha\approx-0.68233$ | Real root of $x^3+x+1$ |
| $x^2+\alpha x+\alpha^2+1$ | Irreducible quadratic factor |
| $A_1,A_2,A_3$ | Coefficients over $(x-\alpha)^k$ |
| $B_k,C_k$ | Coefficients over $(x^2+\alpha x+\alpha^2+1)^k$ |
Additional Notes
- Flag for the reader: this is not a case of a source error (unlike Problem 253, which wrongly claimed impossibility due to a fixable partial-fraction mistake). Here the underlying algebraic number $\alpha$ genuinely has no simpler exact expression than Cardano's radical, so a fully expanded closed-form antiderivative — while it technically exists — is not practical to present by hand at reasonable length.
- The general shape of the final answer is: a rational function of $x$ (from the $A_2,A_3,B_2,C_2,B_3,C_3$ terms after integration) plus $A_1\ln|x-\alpha|$ plus a combination of $\ln(x^2+\alpha x+\alpha^2+1)$ and $\arctan\left(\frac{2x+\alpha}{\sqrt{3\alpha^2+4}}\right)$ terms (from the $B_1,C_1$ term), all with coefficients that are explicit but messy algebraic functions of $\alpha$.
- For a numeric value over a specific interval, direct numerical integration is far more practical than assembling this symbolic closed form.
Result
The integral is elementary in principle but its closed form is impractical to write explicitly by hand; its structure is:
where $R(x)$ is a rational function arising from the repeated-factor terms, $\alpha\approx-0.68233$ is the real root of $x^3+x+1$, and the constants $A_1,B_1,C_1$ come from solving the 9-constant partial-fraction system described above.