What is the integral from 1 to 5 of (4/x-e^{-x}) ?

The integral from 1 to 5 of (4/x-e^{-x}) is 4ln(5)-1/e+1/(e^5)

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integral from 1 to 5 of (4/x-e^{-x})

\int_{1}^{5} (\frac{4}{x} - e^{- x}) d x

4 ln (5) - \frac{1}{e} + \frac{1}{e ^{5}}

Decimal

6.07661 \dots

Solution steps

\int_{1}^{5} \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}^{5} \frac{4}{x} d x - \int_{1}^{5} e^{- x} d x

\int_{1}^{5} \frac{4}{x} d x = 4 ln (5)

\int_{1}^{5} e^{- x} d x = \frac{1}{e} - \frac{1}{e ^{5}}

= 4 ln (5) - (\frac{1}{e} - \frac{1}{e ^{5}})

Simplify

= 4 ln (5) - \frac{1}{e} + \frac{1}{e ^{5}}

What is the integral from 1 to 5 of (4/x-e^{-x}) ?
The integral from 1 to 5 of (4/x-e^{-x}) is 4ln(5)-1/e+1/(e^5)