What is the integral of (e^u)/((4-e^u)^2) ?

The integral of (e^u)/((4-e^u)^2) is 1/(4-e^u)+C

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integral of (e^u)/((4-e^u)^2)

\int \frac{e ^{u}}{( 4 - e ^{u} ) ^{2}} d u

\frac{1}{4 - e ^{u}} + C

Solution steps

\int \frac{e ^{u}}{( 4 - 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 = 4 - e^{u}

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

Simplify

= \frac{1}{4 - e ^{u}}

Add a constant to the solution

= \frac{1}{4 - e ^{u}} + C

\frac{d y}{d x} = y^{2} e^{- x}

\int x \cdot x^{20} d x

t^{2} e^{- 2 t}

x \to 0 lim (e^{0} cos (4) (0))

x \to 3 lim (4 e^{x - 3})

What is the integral of (e^u)/((4-e^u)^2) ?
The integral of (e^u)/((4-e^u)^2) is 1/(4-e^u)+C