What is the integral of (e^{5x}+8e^{-x}) ?

The integral of (e^{5x}+8e^{-x}) is 1/(5e^{-5x)}-8e^{-x}+C

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integral of (e^{5x}+8e^{-x})

\int (e^{5 x} + 8 e^{- x}) d x

\frac{1}{5 e ^{- 5 x}} - 8 e^{- x} + C

Solution steps

\int e^{5 x} + 8 e^{- x} d x

Apply u-substitution

: \int - \frac{1 + 8 u ^{6}}{u ^{6}} d u

= \int - \frac{1 + 8 u ^{6}}{u ^{6}} d u

Take the constant out:

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

= - \int \frac{1 + 8 u ^{6}}{u ^{6}} d u

Expand

\frac{1 + 8 u ^{6}}{u ^{6}} : \frac{1}{u ^{6}} + 8

= - \int \frac{1}{u ^{6}} + 8 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 ^{6}} d u + \int 8 d u)

\int \frac{1}{u ^{6}} d u = - \frac{1}{5 u ^{5}}

\int 8 d u = 8 u

= - (- \frac{1}{5 u ^{5}} + 8 u)

Substitute back

u = e^{- x}

= - (- \frac{1}{5 ( e ^{- x} ) ^{5}} + 8 e^{- x})

Simplify

- (- \frac{1}{5 ( e ^{- x} ) ^{5}} + 8 e^{- x}) : \frac{1}{5 e ^{- 5 x}} - 8 e^{- x}

= \frac{1}{5 e ^{- 5 x}} - 8 e^{- x}

Add a constant to the solution

= \frac{1}{5 e ^{- 5 x}} - 8 e^{- x} + C

\int_{0}^{\frac{π}{2}} (cos (x))^{3} d x

y = \frac{3}{x}, (- 5, - \frac{3}{5})

y = x^{x}, (4, 16)

\frac{d}{d x} (e^{l n (x^{2})})

\frac{d}{d x} (- x^{2} + 4)

What is the integral of (e^{5x}+8e^{-x}) ?
The integral of (e^{5x}+8e^{-x}) is 1/(5e^{-5x)}-8e^{-x}+C