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

The integral of x/(1+e^{-x^2)} is 1/2 (ln|1+e^{-x^2}|+x^2)+C

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 ( 1 + e ^{- u} )} d u

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

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

= \frac{1}{2} \cdot \int - \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 \frac{e ^{- u}}{1 + e ^{- u}} d u + \int 1 d u)

\int \frac{e ^{- u}}{1 + e ^{- u}} d u = - ln 1 + e^{- u}

\int 1 d u = u

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

Substitute back

u = x^{2}

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

Simplify

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

Add a constant to the solution

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

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