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

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

Step 1 · Factor and set up partial fractions

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

Multiplying through: $1=A(x^2-x+1)+(Bx+C)(x+1)$. Setting $x=-1$: $1=3A$, so $A=\tfrac13$. Expanding and matching coefficients of $x^2$ and $x$:

$$ x^2:\ A+B=0 \ \Rightarrow\ B=-\tfrac13, \qquad x:\ -A+B+C=0 \ \Rightarrow\ C=A-B=\tfrac23. $$

Step 2 · Integrate the first term

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

Step 3 · Integrate the second term

With $B=-\tfrac13$, $C=\tfrac23$, write $-\tfrac13x+\tfrac23 = -\tfrac16(2x-1)+\tfrac12$ (since $\dfrac{d}{dx}(x^2-x+1)=2x-1$):

$$ \int \frac{-\frac13x+\frac23}{x^2-x+1}\,dx = -\frac16\int \frac{2x-1}{x^2-x+1}\,dx + \frac12\int \frac{dx}{x^2-x+1}. $$

The first piece is $-\tfrac16\ln(x^2-x+1)$. For the second, completing the square, $x^2-x+1=(x-\tfrac12)^2+\tfrac34$, and using the standard arctangent formula:

$$ \frac12\int \frac{dx}{x^2-x+1} = \frac12\cdot\frac{2}{\sqrt3}\arctan\left(\frac{2x-1}{\sqrt3}\right) = \frac{1}{\sqrt3}\arctan\left(\frac{2x-1}{\sqrt3}\right). $$

Step 4 · Combine

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

Combining the logarithms:

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

Summary Table

Additional Notes

• The domain excludes $x=-1$, the real root of $x^3+1$.
• The result can be checked by differentiating $\tfrac16\ln\!\left(\tfrac{(x+1)^2}{x^2-x+1}\right)+\tfrac1{\sqrt3}\arctan\!\left(\tfrac{2x-1}{\sqrt3}\right)$: combining both derivatives over the common denominator $3(x+1)(x^2-x+1)=3(x^3+1)$ gives numerator $(x^2-x+1)+(x+1)(2-x)=3$, so the whole expression reduces exactly to $1/(x^3+1)$, matching the integrand.