What is the integral of (-5e^{-5x})/((e^{-5x)+3)^3} ?

The integral of (-5e^{-5x})/((e^{-5x)+3)^3} is -1/(2(e^{-5x)+3)^2}+C

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integral of (-5e^{-5x})/((e^{-5x)+3)^3}

\int \frac{- 5 e ^{- 5 x}}{( e ^{- 5 x} + 3 ) ^{3}} d x

- \frac{1}{2 ( e ^{- 5 x} + 3 ) ^{2}} + C

Solution steps

\int \frac{- 5 e ^{- 5 x}}{( e ^{- 5 x} + 3 ) ^{3}} d x

Take the constant out:

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

= - 5 \cdot \int \frac{e ^{- 5 x}}{( e ^{- 5 x} + 3 ) ^{3}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{5 ( e ^{u} + 3 ) ^{3}} d u

= - 5 \cdot \int - \frac{e ^{u}}{5 ( e ^{u} + 3 ) ^{3}} d u

Take the constant out:

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

= - 5 (- \frac{1}{5} \cdot \int \frac{e ^{u}}{( e ^{u} + 3 ) ^{3}} d u)

Apply u-substitution

: \int \frac{1}{v ^{3}} d v

= - 5 (- \frac{1}{5} \cdot \int \frac{1}{v ^{3}} d v)

Apply the Power Rule

: - \frac{1}{2 v ^{2}}

= - 5 (- \frac{1}{5} (- \frac{1}{2 v ^{2}}))

Substitute back

= - 5 (- \frac{1}{5} (- \frac{1}{2 ( e ^{- 5 x} + 3 ) ^{2}}))

Simplify

- 5 (- \frac{1}{5} (- \frac{1}{2 ( e ^{- 5 x} + 3 ) ^{2}})) : - \frac{1}{2 ( e ^{- 5 x} + 3 ) ^{2}}

= - \frac{1}{2 ( e ^{- 5 x} + 3 ) ^{2}}

Add a constant to the solution

= - \frac{1}{2 ( e ^{- 5 x} + 3 ) ^{2}} + C

s \to 2 - lim (s + ∣ 3 s - 6 ∣)

\frac{d}{d x} (\frac{sin ( x ) + cos ( x )}{sin ( x )})

\frac{a x ^{2}}{ln ( \frac{x}{4} )} + \frac{b x ^{4}}{x - 4}

\frac{\partial}{\partial x} (3 x^{2} y^{2} - 1)

x^{2} - 1

What is the integral of (-5e^{-5x})/((e^{-5x)+3)^3} ?
The integral of (-5e^{-5x})/((e^{-5x)+3)^3} is -1/(2(e^{-5x)+3)^2}+C