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

The integral of (e^x)/(sqrt(1+\sqrt{1+e^x))} is 4(1/3 (1+sqrt(1+e^x))^{3/2}-sqrt(1+\sqrt{1+e^x)})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

4 (\frac{1}{3} (1 + 1 + e^{x})^{\frac{3}{2}} - 1 + 1 + e^{x}) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Apply u-substitution

: \int 4 (w^{2} - 1) d w

= \int 4 (w^{2} - 1) d w

Take the constant out:

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

= 4 \cdot \int w^{2} - 1 d w

Apply the Sum Rule:

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

= 4 (\int w^{2} d w - \int 1 d w)

\int w^{2} d w = \frac{w ^{3}}{3}

\int 1 d w = w

= 4 (\frac{w ^{3}}{3} - w)

Substitute back

= 4 \frac{( 1 + 1 + e ^{x} ) ^{3}}{3} - 1 + 1 + e^{x}

Simplify

4 \frac{( 1 + 1 + e ^{x} ) ^{3}}{3} - 1 + 1 + e^{x} : 4 (\frac{1}{3} (1 + 1 + e^{x})^{\frac{3}{2}} - 1 + 1 + e^{x})

= 4 (\frac{1}{3} (1 + 1 + e^{x})^{\frac{3}{2}} - 1 + 1 + e^{x})

Add a constant to the solution

= 4 (\frac{1}{3} (1 + 1 + e^{x})^{\frac{3}{2}} - 1 + 1 + e^{x}) + C

\frac{d}{d x} (e^{- x} cos (x) + e^{- x} sin (x))

ln (3 x + 5)

f (x) = x^{2} - 4 x, at x = - 2

\frac{d}{d x} ((3 x^{2} + 2 x) (x^{4} - 3 x + 1))

y^{^{'''}} + y^{^{'}} = sec (x)

What is the integral of (e^x)/(sqrt(1+\sqrt{1+e^x))} ?
The integral of (e^x)/(sqrt(1+\sqrt{1+e^x))} is 4(1/3 (1+sqrt(1+e^x))^{3/2}-sqrt(1+\sqrt{1+e^x)})+C