What is the integral from 1 to 2 of (9/x-e^{-x}) ?

The integral from 1 to 2 of (9/x-e^{-x}) is 9ln(2)-1/e+1/(e^2)

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integral from 1 to 2 of (9/x-e^{-x})

\int_{1}^{2} (\frac{9}{x} - e^{- x}) d x

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

Decimal

6.00578 \dots

Solution steps

\int_{1}^{2} \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}^{2} \frac{9}{x} d x - \int_{1}^{2} e^{- x} d x

\int_{1}^{2} \frac{9}{x} d x = 9 ln (2)

\int_{1}^{2} e^{- x} d x = \frac{1}{e} - \frac{1}{e ^{2}}

= 9 ln (2) - (\frac{1}{e} - \frac{1}{e ^{2}})

Simplify

= 9 ln (2) - \frac{1}{e} + \frac{1}{e ^{2}}

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What is the integral from 1 to 2 of (9/x-e^{-x}) ?
The integral from 1 to 2 of (9/x-e^{-x}) is 9ln(2)-1/e+1/(e^2)