What is the integral of ((e^u))/((1-e^u)^2) ?

The integral of ((e^u))/((1-e^u)^2) is 1/(1-e^u)+C

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integral of ((e^u))/((1-e^u)^2)

\int \frac{( e ^{u} )}{( 1 - e ^{u} ) ^{2}} d u

\frac{1}{1 - e ^{u}} + C

Solution steps

\int \frac{e ^{u}}{( 1 - e ^{u} ) ^{2}} d u

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

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

Substitute back

v = 1 - e^{u}

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

Simplify

= \frac{1}{1 - e ^{u}}

Add a constant to the solution

= \frac{1}{1 - e ^{u}} + C

What is the integral of ((e^u))/((1-e^u)^2) ?
The integral of ((e^u))/((1-e^u)^2) is 1/(1-e^u)+C