$\int_1^2 \frac{x^4-1}{x^6-1} d x=$
- A$\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
- B$\frac{121}{6}$
- C$\sqrt{2}-1$
- D$\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
✓ Correct answer: A
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