Evaluate the integral: I = integral x/(x³-1) dx

Evaluate the integral: $$ I = \int \frac{x}{x^3-1}\,dx. $$

Step 1 · Factor and set up partial fractions

$$ x^3-1 = (x-1)(x^2+x+1), \qquad \frac{x}{(x-1)(x^2+x+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1}. $$

Multiplying through: $x=A(x^2+x+1)+(Bx+C)(x-1)$. Setting $x=1$ gives $A=\tfrac13$; matching coefficients gives $B=-\tfrac13$ and $C=\tfrac13$.

Step 2 · Integrate the linear-factor term

$$ \frac13\int \frac{dx}{x-1} = \frac13\ln|x-1|. $$

Step 3 · Split the quadratic-factor term

Writing $-\tfrac13x+\tfrac13=-\tfrac16(2x+1)+\tfrac12$:

$$ \int \frac{-\frac13x+\frac13}{x^2+x+1}\,dx = -\frac16\int \frac{2x+1}{x^2+x+1}\,dx + \frac12\int \frac{dx}{x^2+x+1}. $$

The first piece is $-\tfrac16\ln(x^2+x+1)$; the second, using the standard result from Problem 189, is $\tfrac{1}{\sqrt3}\arctan\!\left(\tfrac{2x+1}{\sqrt3}\right)$.

Step 4 · Combine

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

Summary Table

Additional Notes

• The domain excludes $x=1$, the only real root of $x^3-1$ (the quadratic factor $x^2+x+1$ has no real roots).
• This is a standard "cyclotomic" factorization pattern ($x^3-1=(x-1)(x^2+x+1)$), and the resulting quadratic-factor integral always splits into a log term (matching the numerator's derivative) plus the arctangent result reused from Problem 189.
• The result can be checked by differentiating $\tfrac13\ln|x-1|-\tfrac16\ln(x^2+x+1)+\tfrac1{\sqrt3}\arctan\!\left(\tfrac{2x+1}{\sqrt3}\right)$: combining the three derivative terms over $3(x-1)(x^2+x+1)=3(x^3-1)$ gives numerator $(x^2+x+1)-(x-1)^2=3x$, reducing the expression to $\tfrac{3x}{3(x^3-1)}=\tfrac{x}{x^3-1}$, matching the integrand.