Evaluate the integral: I = integral (x³+1)/(x (x²+x+1)²)

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

Step 1 · Rewrite the numerator

Since $x(x^2+x+1)=x^3+x^2+x$, we have $x^3+1 = x(x^2+x+1) - (x^2+x-1)$. Dividing through:

$$ \frac{x^3+1}{x(x^2+x+1)^2} = \frac{1}{x^2+x+1} - \frac{x^2+x-1}{x(x^2+x+1)^2}. $$

Step 2 · Partial fractions for the second piece

$$ \frac{x^2+x-1}{x(x^2+x+1)^2} = \frac Ax + \frac{Bx+C}{x^2+x+1} + \frac{Dx+E}{(x^2+x+1)^2}. $$

Multiplying through: $x^2+x-1=A(x^2+x+1)^2+(Bx+C)x(x^2+x+1)+(Dx+E)x$. Setting $x=0$ gives $A=-1$. Matching the remaining coefficients gives $B=1,\ C=1,\ D=2,\ E=2$.

Step 3 · Integrate each term

$$ I = \int\frac{dx}{x^2+x+1} + \int\frac{dx}{x} - \int\frac{x+1}{x^2+x+1}\,dx - \int\frac{2x+2}{(x^2+x+1)^2}\,dx. $$

Using $x^2+x+1=(x+\tfrac12)^2+\tfrac34$: $\displaystyle\int\frac{dx}{x^2+x+1}=\frac{2}{\sqrt3}\arctan\left(\frac{2x+1}{\sqrt3}\right)$.

Since $x+1=\tfrac12(2x+1)+\tfrac12$: $\displaystyle\int\frac{x+1}{x^2+x+1}\,dx = \frac12\ln\left(x^2+x+1\right) + \frac1{\sqrt3}\arctan\left(\frac{2x+1}{\sqrt3}\right)$.

Since $2x+2=(2x+1)+1$: $\displaystyle\int\frac{2x+2}{(x^2+x+1)^2}\,dx = -\frac1{x^2+x+1} + \frac{2x+1}{3(x^2+x+1)} + \frac{4}{3\sqrt3}\arctan\left(\frac{2x+1}{\sqrt3}\right)$, using $\int\frac{2x+1}{(x^2+x+1)^2}dx=-\frac1{x^2+x+1}$ and the reduction formula for $\int\frac{dx}{(x^2+x+1)^2}$.

Step 4 · Combine

Collecting the arctangent coefficients: $\tfrac{2}{\sqrt3}-\tfrac1{\sqrt3}-\tfrac4{3\sqrt3}=-\tfrac1{3\sqrt3}=-\tfrac{\sqrt3}9$. Collecting the rational terms: $\tfrac1{x^2+x+1}-\tfrac{2x+1}{3(x^2+x+1)}=\tfrac{2-2x}{3(x^2+x+1)}$.

$$ I = \ln|x| - \frac12\ln\left(x^2+x+1\right) - \frac{\sqrt3}9\arctan\left(\frac{2x+1}{\sqrt3}\right) + \frac{2-2x}{3\left(x^2+x+1\right)} + C. $$

Summary Table

Additional Notes

• The domain excludes $x=0$, the only real root of the denominator ($x^2+x+1$ has no real roots).
• The result was verified by differentiating the boxed expression and combining all four terms over the common denominator $6x(x^2+x+1)^2$: the numerator simplifies to exactly $6(x^3+1)$, matching the original integrand.