What is the integral of (e^{3x}+6e^{-x}) ?

The integral of (e^{3x}+6e^{-x}) is 1/(3e^{-3x)}-6e^{-x}+C

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integral of (e^{3x}+6e^{-x})

\int (e^{3 x} + 6 e^{- x}) d x

\frac{1}{3 e ^{- 3 x}} - 6 e^{- x} + C

Solution steps

\int e^{3 x} + 6 e^{- x} d x

Apply u-substitution

: \int - \frac{1 + 6 u ^{4}}{u ^{4}} d u

= \int - \frac{1 + 6 u ^{4}}{u ^{4}} d u

Take the constant out:

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

= - \int \frac{1 + 6 u ^{4}}{u ^{4}} d u

Expand

\frac{1 + 6 u ^{4}}{u ^{4}} : \frac{1}{u ^{4}} + 6

= - \int \frac{1}{u ^{4}} + 6 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 ^{4}} d u + \int 6 d u)

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

\int 6 d u = 6 u

= - (- \frac{1}{3 u ^{3}} + 6 u)

Substitute back

u = e^{- x}

= - (- \frac{1}{3 ( e ^{- x} ) ^{3}} + 6 e^{- x})

Simplify

- (- \frac{1}{3 ( e ^{- x} ) ^{3}} + 6 e^{- x}) : \frac{1}{3 e ^{- 3 x}} - 6 e^{- x}

= \frac{1}{3 e ^{- 3 x}} - 6 e^{- x}

Add a constant to the solution

= \frac{1}{3 e ^{- 3 x}} - 6 e^{- x} + C

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What is the integral of (e^{3x}+6e^{-x}) ?
The integral of (e^{3x}+6e^{-x}) is 1/(3e^{-3x)}-6e^{-x}+C