∫ e^x≤ft((1-x)/(1+x^2))^2dx=

The correct answer is Option (1) → \(\displaystyle\frac{e^x}{1+x^2}+C\)

Expanding the square and splitting the fraction:

    \( \displaystyle \left(\frac{1-x}{1+x^2}\right)^2 = \frac{(1+x^2) - 2x}{(1+x^2)^2} \)

    \( \displaystyle = \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2} \)

With \(f(x) = \frac{1}{1+x^2}\):

    \( \displaystyle f'(x) = -\frac{2x}{(1+x^2)^2} \)

So the integrand is \(e^x\left(f(x) + f'(x)\right)\), and by the standard result \(\int e^x\left(f(x)+f'(x)\right)dx = e^x f(x) + C\):

    \( \displaystyle \int e^x\left(\frac{1-x}{1+x^2}\right)^2dx = \frac{e^x}{1+x^2} + C \)