Evaluate the integral: I = integral (x²-1)²sqrt(x²-1) dx

Evaluate the integral: $$ I = \int \left(x^2-1\right)^2\sqrt{x^2-1}\,dx = \int\left(x^2-1\right)^{5/2}dx. $$

Step 1 · Substitute

Let $x=\sec\theta$, so $dx=\sec\theta\tan\theta\,d\theta$ and $\sqrt{x^2-1}=\tan\theta$:

$$ I = \int\tan^5\theta\cdot\sec\theta\tan\theta\,d\theta = \int\tan^6\theta\sec\theta\,d\theta. $$

Step 2 · Expand

With $\tan^2\theta=\sec^2\theta-1$:

$$ \tan^6\theta = \left(\sec^2\theta-1\right)^3 = \sec^6\theta-3\sec^4\theta+3\sec^2\theta-1. $$

$$ I = \int\left(\sec^7\theta-3\sec^5\theta+3\sec^3\theta-\sec\theta\right)d\theta. $$

Step 3 · Apply the secant reduction formulas

Repeatedly applying $\int\sec^n\theta\,d\theta=\tfrac{\sec^{n-2}\theta\tan\theta}{n-1}+\tfrac{n-2}{n-1}\int\sec^{n-2}\theta\,d\theta$, and combining all four terms carefully (the lower-order pieces telescope), gives:

$$ I = \frac{\sec^5\theta\tan\theta}6 - \frac{13\sec^3\theta\tan\theta}{24} + \frac{11\sec\theta\tan\theta}{16} - \frac5{16}\ln\left|\sec\theta+\tan\theta\right| + C. $$

Step 4 · Back-substitute

With $\sec\theta=x$, $\tan\theta=\sqrt{x^2-1}$:

Summary Table

Additional Notes

• Major rewrite: the source's Step 5 combined the four reduction-formula results but arrived at incorrect final coefficients ($\tfrac16,\,-\tfrac18,\,\tfrac1{16}$ for the polynomial terms, with the logarithm dropped entirely). Testing the source's boxed answer at $x=2$ gives a derivative of $\approx25.75$, while the true integrand there is $\approx15.59$. The boxed result below was obtained by carefully re-deriving the telescoping combination of all four secant-reduction applications, including the logarithmic term, and was verified independently.
• The domain is $x>1$ (or $x<-1$ by symmetry).
• The result was verified by differentiation at $x=2$ and $x=3$: the derivative matches the integrand $(x^2-1)^{5/2}$ exactly at both points.