What is the integral of ((e^x+2)e^x)/(sqrt(e^x+1)) ?

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

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

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

Solution steps

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

Expand

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

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

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

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

\int \frac{2 e ^{x}}{e ^{x} + 1} d x = 4 e^{x} + 1

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

Simplify

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

Add a constant to the solution

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

\int \frac{x ^{4} + 6 x ^{2}}{x ( x ^{2} + 4 ) ^{2}} d x

\int x^{2} 16 - x^{3} d x

\int e^{2 x} sin (\frac{x}{2}) d x

\int (x + 2) (x^{2} - 1) d x

\int 11 x^{- 6} d x

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