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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{u ^{3}} d u

= \int \frac{u - 1}{u ^{3}} d u

Expand

\frac{u - 1}{u ^{3}} : \frac{1}{u ^{2}} - \frac{1}{u ^{3}}

= \int \frac{1}{u ^{2}} - \frac{1}{u ^{3}} 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{1}{u ^{2}} d u - \int \frac{1}{u ^{3}} d u

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

= - \frac{1}{u} - (- \frac{1}{2 u ^{2}})

Substitute back

u = e^{x} + 1

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

Simplify

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

Add a constant to the solution

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

\int_{- π}^{π} cos^{2} (9 x) d x

\int \frac{cos ( a x )}{sin ^{5} ( a x )} d x

7 t sin (2 t)

\frac{d}{d x} (x y e^{\frac{- x ^{2}}{2} - \frac{y ^{2}}{2}})

\int r 9 - r^{2} d r

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