Evaluate the integral: I = integral (x^5)/(x^6-x³-2) dx

Evaluate the integral: $$ I = \int \frac{x^5}{x^6-x^3-2}\,dx. $$

Step 1 · Substitute $u=x^3$

Let $u=x^3$, $du=3x^2\,dx$. Writing $x^5\,dx=x^3\cdot x^2\,dx=u\cdot\dfrac{du}3$:

$$ I = \frac13\int \frac{u}{u^2-u-2}\,du. $$

Step 2 · Factor and decompose

$$ u^2-u-2=(u-2)(u+1), \qquad \frac{u}{(u-2)(u+1)} = \frac{A}{u-2}+\frac{B}{u+1}. $$

Multiplying through: $u=A(u+1)+B(u-2)$. Setting $u=2$: $2=3A$, so $A=\tfrac23$. Setting $u=-1$: $-1=-3B$, so $B=\tfrac13$.

Step 3 · Integrate

$$ I = \frac13\int \left(\frac{2/3}{u-2}+\frac{1/3}{u+1}\right) du = \frac29\ln|u-2| + \frac19\ln|u+1| + C. $$

Step 4 · Substitute back

With $u=x^3$:

$$ I = \frac29\ln\left|x^3-2\right| + \frac19\ln\left|x^3+1\right| + C. $$

Summary Table

Additional Notes

• The domain excludes $x^3=2$ and $x^3=-1$, i.e. $x=\sqrt[3]2$ and $x=-1$.
• This problem is identical to Problem 225 elsewhere in this collection, and both were solved with the same substitution and give the same verified answer.
• The result can be checked by differentiating $\tfrac29\ln|x^3-2|+\tfrac19\ln|x^3+1|$: this gives $\tfrac{2}{9}\cdot\tfrac{3x^2}{x^3-2}+\tfrac19\cdot\tfrac{3x^2}{x^3+1}=\tfrac{2x^2}{3(x^3-2)}+\tfrac{x^2}{3(x^3+1)}$, which combines over the common denominator $3(x^3-2)(x^3+1)=3(x^6-x^3-2)$ to give numerator $2x^2(x^3+1)+x^2(x^3-2)=3x^5$, so the whole expression reduces exactly to $x^5/(x^6-x^3-2)$.