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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

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

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

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

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

\int 1 d u = u

= - (u + 2 ln u + 1 + 1 - 2 u + 1 - 2) + u

Substitute back

u = e^{x}

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

Simplify

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

Add a constant to the solution

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

\int \frac{( 2 x + 3 ) ^{2}}{x} d x

\frac{d}{d x} (2 sin (x))

\frac{2}{1 - t}, at t = - 2

\frac{d}{d x} (2 x ln (2 x) + x)

\frac{d y}{d x} - \frac{2 x}{2 - x ^{2}} \cdot y = e^{x}, y (2) = 4

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