MCQ
$\int\text{e}^\text{x}(\frac{1-\text{x}}{1+\text{x}^2})^2\text{dx}$ is equal to:
  • $\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{c}$
  • B
    $-\frac{-\text{e}^\text{x}}{1+\text{x}^2}+\text{c}$
  • C
    $\frac{\text{e}^\text{x}}{(1+\text{x}^2)^2}+\text{c}$
  • D
    $\frac{-\text{e}^\text{x}}{(1+\text{x}^2)^2}+\text{c}$

Answer

Correct option: A.
$\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{c}$
$\frac{\text{e}^\text{x}}{1+\text{x}^2}+\text{c}$

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