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$\frac{-\text{e}^\text{x}}{(1+\text{x}^2)^2}+\text{c}$