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

• A

  $\frac{\left(\operatorname{Tan}^{-1} x\right)^2}{4}-\frac{x \operatorname{Tan}^{-1} x}{2\left(1+x^2\right)}+\frac{1-x^2}{4\left(1+x^2\right)}+c$
• Correct answer
  B

  $\frac{\left(\operatorname{Tan}^{-1} x\right)^2}{4}-\frac{4 x \operatorname{Tan}^{-1} x+1-x^2}{8\left(1+x^2\right)}+\mathrm{c}$
• C

  $\frac{\left(\operatorname{Tan}^{-1} x\right)^2}{4}-\frac{x \operatorname{Tan}^{-1} x}{\left(1+x^2\right)}-\frac{1-x^2}{4\left(1+x^2\right)}+\mathrm{c}$
• D

  $\frac{(\tan x)^2}{4}+\frac{4 x \operatorname{Tan}^{-1} x-1+x^2}{4\left(1+x^2\right)}+\mathrm{c}$

Hence, the correct answer is option: 'B'.