What is the integral from 1 to 8 of (4/x-e^{-x}) ?

The integral from 1 to 8 of (4/x-e^{-x}) is 12ln(2)-1/e+1/(e^8)

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integral from 1 to 8 of (4/x-e^{-x})

\int_{1}^{8} (\frac{4}{x} - e^{- x}) d x

12 ln (2) - \frac{1}{e} + \frac{1}{e ^{8}}

Decimal

7.95022 \dots

Solution steps

\int_{1}^{8} \frac{4}{x} - e^{- 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_{1}^{8} \frac{4}{x} d x - \int_{1}^{8} e^{- x} d x

\int_{1}^{8} \frac{4}{x} d x = 12 ln (2)

\int_{1}^{8} e^{- x} d x = \frac{1}{e} - \frac{1}{e ^{8}}

= 12 ln (2) - (\frac{1}{e} - \frac{1}{e ^{8}})

Simplify

= 12 ln (2) - \frac{1}{e} + \frac{1}{e ^{8}}

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What is the integral from 1 to 8 of (4/x-e^{-x}) ?
The integral from 1 to 8 of (4/x-e^{-x}) is 12ln(2)-1/e+1/(e^8)