Evaluate the integral: I = integral (x^5)/(x³-1) dx

Evaluate the integral: $$ I = \int \frac{x^5}{x^3-1}\,dx. $$

Step 1 · Split the numerator

Since $x^5=x^2\cdot x^3=x^2\left[(x^3-1)+1\right]=x^2(x^3-1)+x^2$:

$$ \frac{x^5}{x^3-1} = x^2 + \frac{x^2}{x^3-1}. $$

Step 2 · Integrate the polynomial piece

$$ \int x^2\,dx = \frac{x^3}{3}. $$

Step 3 · Integrate the remaining piece directly

Since $\dfrac{d}{dx}(x^3-1)=3x^2$, the numerator $x^2$ is already (up to a constant) the derivative of the denominator:

$$ \int \frac{x^2}{x^3-1}\,dx = \frac13\int \frac{3x^2}{x^3-1}\,dx = \frac13\ln\left|x^3-1\right|. $$

Step 4 · Combine

$$ I = \frac{x^3}{3} + \frac13\ln\left|x^3-1\right| + C. $$

Summary Table

Additional Notes

• Correction: the source set up partial fractions for $x^2/[(x-1)(x^2+x+1)]$ but solved the resulting linear system incorrectly, claiming $A=B=C=\tfrac12$; the correct values (found by matching coefficients directly against $x^2=A(x^2+x+1)+(Bx+C)(x-1)$) are $A=\tfrac13$, $B=\tfrac23$, $C=\tfrac13$. However, none of this partial-fraction machinery is actually necessary here: since the numerator $x^2$ is exactly $\tfrac13$ of the derivative of the denominator $x^3-1$, the whole piece integrates directly to $\tfrac13\ln|x^3-1|$ — which, reassuringly, is exactly what the (correct) partial fraction route also gives, since $\tfrac13\ln|x-1|+\tfrac13\ln(x^2+x+1)=\tfrac13\ln|x^3-1|$.
• The domain excludes $x=1$, the real root of $x^3-1$.
• The result can be checked by differentiating $\tfrac{x^3}3+\tfrac13\ln|x^3-1|$: this gives $x^2+\tfrac{x^2}{x^3-1}=\tfrac{x^2(x^3-1)+x^2}{x^3-1}=\tfrac{x^5-x^2+x^2}{x^3-1}=\tfrac{x^5}{x^3-1}$, matching the integrand.