What is the integral of (e^{9x}+7e^{-x}) ?

The integral of (e^{9x}+7e^{-x}) is 1/(9e^{-9x)}-7e^{-x}+C

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integral of (e^{9x}+7e^{-x})

\int (e^{9 x} + 7 e^{- x}) d x

\frac{1}{9 e ^{- 9 x}} - 7 e^{- x} + C

Solution steps

\int e^{9 x} + 7 e^{- x} d x

Apply u-substitution

: \int - \frac{1 + 7 u ^{10}}{u ^{10}} d u

= \int - \frac{1 + 7 u ^{10}}{u ^{10}} d u

Take the constant out:

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

= - \int \frac{1 + 7 u ^{10}}{u ^{10}} d u

Expand

\frac{1 + 7 u ^{10}}{u ^{10}} : \frac{1}{u ^{10}} + 7

= - \int \frac{1}{u ^{10}} + 7 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int \frac{1}{u ^{10}} d u + \int 7 d u)

\int \frac{1}{u ^{10}} d u = - \frac{1}{9 u ^{9}}

\int 7 d u = 7 u

= - (- \frac{1}{9 u ^{9}} + 7 u)

Substitute back

u = e^{- x}

= - (- \frac{1}{9 ( e ^{- x} ) ^{9}} + 7 e^{- x})

Simplify

- (- \frac{1}{9 ( e ^{- x} ) ^{9}} + 7 e^{- x}) : \frac{1}{9 e ^{- 9 x}} - 7 e^{- x}

= \frac{1}{9 e ^{- 9 x}} - 7 e^{- x}

Add a constant to the solution

= \frac{1}{9 e ^{- 9 x}} - 7 e^{- x} + C

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What is the integral of (e^{9x}+7e^{-x}) ?
The integral of (e^{9x}+7e^{-x}) is 1/(9e^{-9x)}-7e^{-x}+C