$\int \frac{(\tan^{-1} x)^3}{1 + x^2} \, dx = $
- A$\frac{(\tan^{-1} x)^4}{4} + c$
- B$\frac{(\tan^{-1} x)^4}{4} + c$
- C$2 \tan^{-1} x + c$
- D$2 (\tan^{-1} x)^2 + c$
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