What is the integral of (e^{-x})/((9e^{-2x)+1)^{3/2}} ?

The integral of (e^{-x})/((9e^{-2x)+1)^{3/2}} is -(e^{-x})/((1+9e^{-2x))^{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^{-x})/((9e^{-2x)+1)^{3/2}}

\int \frac{e ^{- x}}{( 9 e ^{- 2 x} + 1 ) ^{\frac{3}{2}}} d x

- \frac{e ^{- x}}{( 1 + 9 e ^{- 2 x} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{e ^{- x}}{( 9 e ^{- 2 x} + 1 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int - \frac{1}{( 9 u ^{2} + 1 ) ^{\frac{3}{2}}} d u

= \int - \frac{1}{( 9 u ^{2} + 1 ) ^{\frac{3}{2}}} d u

Take the constant out:

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

= - \int \frac{1}{( 9 u ^{2} + 1 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int \frac{1}{3 sec ( v )} d v

= - \int \frac{1}{3 sec ( v )} d v

Take the constant out:

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

= - \frac{1}{3} \cdot \int \frac{1}{sec ( v )} d v

Rewrite using trig identities

= - \frac{1}{3} \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= - \frac{1}{3} sin (v)

Substitute back

= - \frac{1}{3} sin (arctan (3 e^{- x}))

Simplify

- \frac{1}{3} sin (arctan (3 e^{- x})) : - \frac{e ^{- x}}{( 1 + 9 e ^{- 2 x} ) ^{\frac{1}{2}}}

= - \frac{e ^{- x}}{( 1 + 9 e ^{- 2 x} ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{e ^{- x}}{( 1 + 9 e ^{- 2 x} ) ^{\frac{1}{2}}} + C

1 - x^{2}

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{2}} - y^{2})

\int sin (13 x) d x

\int_{0}^{1} ln (4 x) d x

\int - \frac{4}{x} d x

What is the integral of (e^{-x})/((9e^{-2x)+1)^{3/2}} ?
The integral of (e^{-x})/((9e^{-2x)+1)^{3/2}} is -(e^{-x})/((1+9e^{-2x))^{1/2}}+C