$\int \frac{x^2 \operatorname{Tan}^{-1} x}{(1+x^2)^2} dx =$

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

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