$$\int {{{{x^3}dx} \over {1 + {x^8}}} = } $$
- A$$4{\tan ^{ - 1}}{x^3} + c$$
- B$${1 \over 4}{\tan ^{ - 1}}{x^4} + c$$
- C$$x + 4{\tan ^{ - 1}}{x^4} + c$$
- D$${x^2} + {1 \over 4}{\tan ^{ - 1}}{x^4} + c$$
✓ Correct answer: B
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