What is the integral from 0 to 1 of x/(1+e^{-x^2)} ?

The integral from 0 to 1 of x/(1+e^{-x^2)} is 1/2 (-ln(2)+ln(1+1/e)+1)

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integral from 0 to 1 of x/(1+e^{-x^2)}

\int_{0}^{1} \frac{x}{1 + e ^{- x^{2}}} d x

\frac{1}{2} (- ln (2) + ln (1 + \frac{1}{e}) + 1)

Decimal

0.31005 \dots

Solution steps

\int_{0}^{1} \frac{x}{1 + e ^{- x^{2}}} d x

Apply u-substitution

= \int_{0}^{1} \frac{1}{2 ( 1 + e ^{- u} )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{1 + e ^{- u}} d u

\frac{1}{1 + e ^{- u}} = - \frac{e ^{- u}}{1 + e ^{- u}} + 1

= \frac{1}{2} \cdot \int_{0}^{1} - \frac{e ^{- u}}{1 + e ^{- u}} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int_{0}^{1} \frac{e ^{- u}}{1 + e ^{- u}} d u + \int_{0}^{1} 1 d u)

\int_{0}^{1} \frac{e ^{- u}}{1 + e ^{- u}} d u = ln (2) - ln (1 + \frac{1}{e})

\int_{0}^{1} 1 d u = 1

= \frac{1}{2} (- (ln (2) - ln (1 + \frac{1}{e})) + 1)

Simplify

= \frac{1}{2} (- ln (2) + ln (1 + \frac{1}{e}) + 1)

What is the integral from 0 to 1 of x/(1+e^{-x^2)} ?
The integral from 0 to 1 of x/(1+e^{-x^2)} is 1/2 (-ln(2)+ln(1+1/e)+1)