Evaluate the integral: I = integral (x²)/(sqrt(x^6-1)) dx

Evaluate the integral: $$ I = \int \frac{x^2}{\sqrt{x^6-1}}\,dx. $$

Explanation of Method

The substitution $u=x^3$ turns $x^2\,dx$ into $\tfrac13du$ and $x^6-1$ into $u^2-1$, reducing the integral directly to the standard logarithmic form $\int du/\sqrt{u^2-1}$.

Substitute $u=x^3$.
Apply the standard formula.
Back-substitute.

Step 1 · Substitute

Let $u=x^3$, so $du=3x^2\,dx$ and $x^6-1=u^2-1$:

$$ I = \frac13\int\frac{du}{\sqrt{u^2-1}}. $$

Step 2 · Apply the standard formula

$$ \int\frac{du}{\sqrt{u^2-1}} = \ln\left|u+\sqrt{u^2-1}\right| + C. $$

Step 3 · Back-substitute

$$ I = \frac13\ln\left|x^3+\sqrt{x^6-1}\right| + C. $$

Summary Table

Substitution	Effect
$u=x^3$	$x^6-1\to u^2-1$, $x^2dx\to du/3$

Additional Notes

The domain requires $x^6>1$, i.e. $|x|>1$.
The result was verified by direct differentiation: $\tfrac{d}{dx}\left[\tfrac13\ln|x^3+\sqrt{x^6-1}|\right]$ simplifies exactly to $x^2/\sqrt{x^6-1}$, confirmed numerically at $x=2$.

Result

The solution to the integral is:

$$ I = \frac13\ln\left|x^3+\sqrt{x^6-1}\right| + C. $$