What is the integral of e^{-x}(9+e^{-x}) ?

The integral of e^{-x}(9+e^{-x}) is -((9+e^{-x})^2)/2+C

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integral of e^{-x}(9+e^{-x})

\int e^{- x} (9 + e^{- x}) d x

- \frac{( 9 + e ^{- x} ) ^{2}}{2} + C

Solution steps

\int e^{- x} (9 + e^{- x}) d x

Apply u-substitution

: \int - u d u

= \int - u d u

Take the constant out:

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

= - \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{u ^{2}}{2}

Substitute back

u = 9 + e^{- x}

= - \frac{( 9 + e ^{- x} ) ^{2}}{2}

Add a constant to the solution

= - \frac{( 9 + e ^{- x} ) ^{2}}{2} + C

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What is the integral of e^{-x}(9+e^{-x}) ?
The integral of e^{-x}(9+e^{-x}) is -((9+e^{-x})^2)/2+C