What is the integral of (e^u)/((9-e^u)^2) ?

The integral of (e^u)/((9-e^u)^2) is 1/(9-e^u)+C

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integral of (e^u)/((9-e^u)^2)

\int \frac{e ^{u}}{( 9 - e ^{u} ) ^{2}} d u

\frac{1}{9 - e ^{u}} + C

Solution steps

\int \frac{e ^{u}}{( 9 - e ^{u} ) ^{2}} d u

Apply u-substitution

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

= \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}

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

Substitute back

v = 9 - e^{u}

= - (- \frac{1}{9 - e ^{u}})

Simplify

= \frac{1}{9 - e ^{u}}

Add a constant to the solution

= \frac{1}{9 - e ^{u}} + C

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What is the integral of (e^u)/((9-e^u)^2) ?
The integral of (e^u)/((9-e^u)^2) is 1/(9-e^u)+C