What is the integral of (e^x)/(1+sqrt(e^x)) ?

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

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integral of (e^x)/(1+sqrt(e^x))

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{1 + u} d u

= \int \frac{1}{1 + u} d u

Apply u-substitution

: \int \frac{2 ( v - 1 )}{v} d v

= \int \frac{2 ( v - 1 )}{v} d v

Take the constant out:

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

= 2 \cdot \int \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= 2 \cdot \int 1 - \frac{1}{v} d v

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 1 d v - \int \frac{1}{v} d v)

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 (v - ln ∣ v ∣)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

y = \frac{x ^{2}}{x - 4}

f (x) = 1 + x^{2}

\frac{d y}{d x} = x - 5

\int 3 3 x + 1 d x

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