What is the integral from 1 to 6 of (9/x-e^{-x}) ?

The integral from 1 to 6 of (9/x-e^{-x}) is 9ln(6)-1/e+1/(e^6)

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integral from 1 to 6 of (9/x-e^{-x})

\int_{1}^{6} (\frac{9}{x} - e^{- x}) d x

9 ln (6) - \frac{1}{e} + \frac{1}{e ^{6}}

Decimal Notation

15.76043 \dots

Solution steps

\int_{1}^{6} \frac{9}{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}^{6} \frac{9}{x} d x - \int_{1}^{6} e^{- x} d x

\int_{1}^{6} \frac{9}{x} d x = 9 ln (6)

\int_{1}^{6} e^{- x} d x = \frac{1}{e} - \frac{1}{e ^{6}}

= 9 ln (6) - (\frac{1}{e} - \frac{1}{e ^{6}})

Simplify

= 9 ln (6) - \frac{1}{e} + \frac{1}{e ^{6}}

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What is the integral from 1 to 6 of (9/x-e^{-x}) ?
The integral from 1 to 6 of (9/x-e^{-x}) is 9ln(6)-1/e+1/(e^6)