$\int \frac{x^3 \tan^{-1} x^4}{1+x^8} dx =$

  • A
    $\frac{(\tan^{-1}(x^4))^2}{8} + c$
  • B
    $\frac{(\tan^{-1}(x^4))^3}{3} + c$
  • C
    $\frac{(\tan^{-1}(x^4))^2}{4} + c$
  • D
    $\frac{(\tan^{-1}(x^4))^2}{2} + c$

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