∫ e^x≤ft((1-x)/(1+x^2))^2 dx is equal to

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

Expand the square and split it:

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

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

With \(f(x) = \frac{1}{1+x^2}\), we have \(f'(x) = -\frac{2x}{(1+x^2)^2}\), so the integrand is \(e^x\left[f(x) + f'(x)\right]\). By the standard result \(\int e^x\left[f(x) + f'(x)\right]dx = e^x f(x) + C\):

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