Question

Solve: $\int x^2\sqrt{x^6-1}dx$

Answer 1

$t=x^3⟹dt=3x^{2}dx$

$\int x^2\sqrt{x^6-1}dx=\frac{1}{3}\int \sqrt{t^2-1}dt=\frac{t}{6}\sqrt{t^2-1}+\frac{1}{6}\ln (t+\sqrt{t^2-1})+k=\frac{x^3}{6}\sqrt{x^6-1}+\frac{1}{6}\ln (x^3+\sqrt{x^6-1})+k$