What is the integral from 0 to ln(3) of e^x-e^{-4x} ?

The integral from 0 to ln(3) of e^x-e^{-4x} is 142/81

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integral from 0 to ln(3) of e^x-e^{-4x}

\int_{0}^{l n (3)} e^{x} - e^{- 4 x} d x

\frac{142}{81}

Decimal Notation

1.75308 \dots

Solution steps

\int_{0}^{l n (3)} e^{x} - e^{- 4 x} d x

Apply the Sum Rule:

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

= \int_{0}^{l n (3)} e^{x} d x - \int_{0}^{l n (3)} e^{- 4 x} d x

\int_{0}^{l n (3)} e^{x} d x = 2

\int_{0}^{l n (3)} e^{- 4 x} d x = \frac{20}{81}

= 2 - \frac{20}{81}

Simplify

2 - \frac{20}{81} : \frac{142}{81}

= \frac{142}{81}

\frac{\partial}{\partial y} (1 + y^{2} sin (2 x))

\int x (1 + x^{2}) d x

\frac{d}{d x} (x^{3} - 12 x^{2} + 45 x)

x \to 2 lim (\frac{- 2}{0})

11 x - 4 y x^{2} + 1 y^{^{'}} = 0, y (0) = 2

What is the integral from 0 to ln(3) of e^x-e^{-4x} ?
The integral from 0 to ln(3) of e^x-e^{-4x} is 142/81