What is the integral of (e^x)/((1-e^x)^5) ?

The integral of (e^x)/((1-e^x)^5) is 1/(4(1-e^x)^4)+C

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integral of (e^x)/((1-e^x)^5)

\int \frac{e ^{x}}{( 1 - e ^{x} ) ^{5}} d x

\frac{1}{4 ( 1 - e ^{x} ) ^{4}} + C

Solution steps

\int \frac{e ^{x}}{( 1 - e ^{x} ) ^{5}} d x

Apply u-substitution

: \int - \frac{1}{u ^{5}} d u

= \int - \frac{1}{u ^{5}} d u

Take the constant out:

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

= - \int \frac{1}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= - (- \frac{1}{4 u ^{4}})

Substitute back

u = 1 - e^{x}

= - (- \frac{1}{4 ( 1 - e ^{x} ) ^{4}})

Simplify

= \frac{1}{4 ( 1 - e ^{x} ) ^{4}}

Add a constant to the solution

= \frac{1}{4 ( 1 - e ^{x} ) ^{4}} + C

\int \frac{x + 2}{x - 2} d x

f (x) = \frac{1}{x + 8}, at x = 5

x \to 4 - lim (\frac{( 1 - x )}{4 - x})

cos (6 x) (\frac{d y}{d x}) + 6 (tan (6 x)) y = 5 sec (6 x)

t \to 0 lim (cos (\frac{1}{t}))

What is the integral of (e^x)/((1-e^x)^5) ?
The integral of (e^x)/((1-e^x)^5) is 1/(4(1-e^x)^4)+C