What is the integral of (-5e^{-5x})/((e^{-5x)+5)^6} ?

The integral of (-5e^{-5x})/((e^{-5x)+5)^6} is -1/(5(e^{-5x)+5)^5}+C

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integral of (-5e^{-5x})/((e^{-5x)+5)^6}

\int \frac{- 5 e ^{- 5 x}}{( e ^{- 5 x} + 5 ) ^{6}} d x

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

Solution steps

\int \frac{- 5 e ^{- 5 x}}{( e ^{- 5 x} + 5 ) ^{6}} 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} + 5 ) ^{6}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{5 ( e ^{u} + 5 ) ^{6}} d u

= - 5 \cdot \int - \frac{e ^{u}}{5 ( e ^{u} + 5 ) ^{6}} 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} + 5 ) ^{6}} d u)

Apply u-substitution

: \int \frac{1}{v ^{6}} d v

= - 5 (- \frac{1}{5} \cdot \int \frac{1}{v ^{6}} d v)

Apply the Power Rule

: - \frac{1}{5 v ^{5}}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (-5e^{-5x})/((e^{-5x)+5)^6} ?
The integral of (-5e^{-5x})/((e^{-5x)+5)^6} is -1/(5(e^{-5x)+5)^5}+C