The value of ∫_0^1 e^x≤ft(1/(1+x)-1/((1+x)^2))dx is

The correct answer is Option (4) → \(\displaystyle\frac{e}{2}-1\)

Let \(f(x) = \displaystyle\frac{1}{1+x}\). Then

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

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

    \( \displaystyle \int_{0}^{1}e^{x}\left(\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\right)dx = \left[\frac{e^{x}}{1+x}\right]_{0}^{1} \)

    \( \displaystyle = \frac{e}{2} - \frac{1}{1} = \frac{e}{2} - 1 \)

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{e}{2}\) — Incorrect. It leaves out the value at the lower limit, \(\frac{e^{0}}{1+0} = 1\).

• Option (2) \(e-1\) — Incorrect. This is \(\int_{0}^{1}e^{x}\,dx\); the bracket multiplying \(e^{x}\) has been ignored.

• Option (3) \(\displaystyle\frac{e-1}{2}\) — Incorrect. At \(x = 0\) the antiderivative \(\frac{e^{x}}{1+x}\) is 1, not \(\frac{1}{2}\).