What is the integral of (e^{3x}+5e^{-x}) ?

The integral of (e^{3x}+5e^{-x}) is 1/(3e^{-3x)}-5e^{-x}+C

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integral of (e^{3x}+5e^{-x})

\int (e^{3 x} + 5 e^{- x}) d x

\frac{1}{3 e ^{- 3 x}} - 5 e^{- x} + C

Solution steps

\int e^{3 x} + 5 e^{- x} d x

Apply u-substitution

: \int - \frac{1 + 5 u ^{4}}{u ^{4}} d u

= \int - \frac{1 + 5 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 + 5 u ^{4}}{u ^{4}} d u

Expand

\frac{1 + 5 u ^{4}}{u ^{4}} : \frac{1}{u ^{4}} + 5

= - \int \frac{1}{u ^{4}} + 5 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 5 d u)

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

\int 5 d u = 5 u

= - (- \frac{1}{3 u ^{3}} + 5 u)

Substitute back

u = e^{- x}

= - (- \frac{1}{3 ( e ^{- x} ) ^{3}} + 5 e^{- x})

Simplify

- (- \frac{1}{3 ( e ^{- x} ) ^{3}} + 5 e^{- x}) : \frac{1}{3 e ^{- 3 x}} - 5 e^{- x}

= \frac{1}{3 e ^{- 3 x}} - 5 e^{- x}

Add a constant to the solution

= \frac{1}{3 e ^{- 3 x}} - 5 e^{- x} + C

\frac{d}{d x} (c_{1} cos (x) + c_{2} sin (x))

\int sin (p) d p

y^{^{''}} - 4 y^{^{'}} + 4 y = 0

\int \frac{x}{7 x ^{2} + 3} d x

\frac{d}{d x} (ln (x^{3} - \frac{1}{x}))

What is the integral of (e^{3x}+5e^{-x}) ?
The integral of (e^{3x}+5e^{-x}) is 1/(3e^{-3x)}-5e^{-x}+C