Evaluate the integral: I = integral dx/((1+x) (1+x²)

Evaluate the integral: $$ I = \int \frac{dx}{(1+x)\left(1+x^2\right)\left(1+x^3\right)}. $$

Step 1 · Factor and identify the repeated root

$$ 1+x^3=(1+x)\left(x^2-x+1\right) \quad\Rightarrow\quad (1+x)(1+x^2)(1+x^3) = (1+x)^2(1+x^2)(x^2-x+1). $$

The factor $(1+x)$ is repeated. A partial-fraction ansatz that treats $1+x$, $1+x^2$, $1+x^3$ as three independent simple factors misses this repetition and cannot succeed — it under-counts the degrees of freedom needed to match the denominator.

Step 2 · Correct partial fraction setup

$$ \frac{1}{(1+x)^2(1+x^2)(x^2-x+1)} = \frac{A}{1+x} + \frac{B}{(1+x)^2} + \frac{Cx+D}{1+x^2} + \frac{Ex+F}{x^2-x+1}. $$

Step 3 · Solve for the coefficients

Setting $x=-1$ isolates $B$: the other terms vanish except $B(1+x^2)(x^2-x+1)$, giving $1=B(2)(3)$, so $B=\tfrac16$.

Setting $x=i$ (a root of $1+x^2=0$) isolates $C,D$: $1=(Ci+D)(1+i)^2(i^2-i+1) = (Ci+D)(2i)(-i) = 2(Ci+D)$, so $Ci+D=\tfrac12$, giving $C=0,\ D=\tfrac12$.

Setting $x=\omega=e^{i\pi/3}$ (a root of $x^2-x+1=0$) isolates $E,F$: working through $(E\omega+F)(1+\omega)^2(1+\omega^2)=1$ and separating real and imaginary parts gives $F=0,\ E=-\tfrac13$.

Matching the $x^5$ coefficient, $A+C+E=0$, then gives $A=\tfrac13$.

Step 4 · Integrate each term

$$ \int\frac{1/3}{1+x}\,dx = \frac13\ln|1+x|, \qquad \int\frac{1/6}{(1+x)^2}\,dx = -\frac1{6(1+x)}, \qquad \int\frac{1/2}{1+x^2}\,dx=\frac12\arctan x. $$

For the last term, split $x=\tfrac12(2x-1)+\tfrac12$:

$$ -\frac13\int\frac{x\,dx}{x^2-x+1} = -\frac16\int\frac{(2x-1)\,dx}{x^2-x+1} - \frac16\int\frac{dx}{x^2-x+1} = -\frac16\ln\left(x^2-x+1\right) - \frac{\sqrt3}{9}\arctan\left(\frac{2x-1}{\sqrt3}\right). $$

Step 5 · Combine

$$ I = \frac13\ln|1+x| - \frac1{6(1+x)} + \frac12\arctan x - \frac16\ln\left(x^2-x+1\right) - \frac{\sqrt3}{9}\arctan\left(\frac{2x-1}{\sqrt3}\right) + C. $$

Summary Table

Additional Notes

• Correction: the source treated $(1+x)$, $(1+x^2)$, $(1+x^3)$ as three independent simple factors, missing that $1+x^3=(1+x)(x^2-x+1)$ shares the root $x=-1$ with $1+x$. This under-determined ansatz produced a contradictory system ($0=1$) at the last step, which the source misread as evidence the integral has no elementary antiderivative. That conclusion is incorrect: every rational function of $x$ has an elementary antiderivative expressible via logarithms and arctangents, since its denominator always factors (over the reals) into linear and irreducible quadratic pieces, and partial fractions always resolves it — the source's substitution $u=1+x$ attempt in Step 7 was already on the right track (it correctly expanded $1+x^2$ and $1+x^3$ in terms of $u$) but was abandoned just before recognizing the shared factor.
• The domain excludes $x=-1$, the only real root of the denominator ($x^2+1$ and $x^2-x+1$ have no real roots).
• The result was verified two ways: symbolically, by confirming all six partial-fraction coefficient equations (matching powers $x^0$ through $x^5$) are satisfied simultaneously, and numerically, by evaluating the derivative of the boxed antiderivative at $x=0$ and $x=1$ and confirming it equals the original integrand at each point ($1$ and $1/8$ respectively).