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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

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

= 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

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

Substitute back

u = x

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

Simplify

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

Add a constant to the solution

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

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