What is the integral of (e^x)/((e^x+1)^9) ?

The integral of (e^x)/((e^x+1)^9) is -1/(8(e^x+1)^8)+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)/((e^x+1)^9)

\int \frac{e ^{x}}{( e ^{x} + 1 ) ^{9}} d x

- \frac{1}{8 ( e ^{x} + 1 ) ^{8}} + C

Solution steps

\int \frac{e ^{x}}{( e ^{x} + 1 ) ^{9}} d x

Apply u-substitution

: \int \frac{1}{( u + 1 ) ^{9}} d u

= \int \frac{1}{( u + 1 ) ^{9}} d u

Apply u-substitution

: \int \frac{1}{v ^{9}} d v

= \int \frac{1}{v ^{9}} d v

Apply the Power Rule

: - \frac{1}{8 v ^{8}}

= - \frac{1}{8 v ^{8}}

Substitute back

= - \frac{1}{8 ( e ^{x} + 1 ) ^{8}}

Add a constant to the solution

= - \frac{1}{8 ( e ^{x} + 1 ) ^{8}} + C

\int e^{3 x} \cdot cos (x) d x

\int 12 e^{2 x} d x

\int (r + 5)^{3} d r

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

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

What is the integral of (e^x)/((e^x+1)^9) ?
The integral of (e^x)/((e^x+1)^9) is -1/(8(e^x+1)^8)+C