What is the integral from 1 to 3 of (6/x-e^{-x}) ?

The integral from 1 to 3 of (6/x-e^{-x}) is 6ln(3)-1/e+1/(e^3)

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integral from 1 to 3 of (6/x-e^{-x})

\int_{1}^{3} (\frac{6}{x} - e^{- x}) d x

6 ln (3) - \frac{1}{e} + \frac{1}{e ^{3}}

Decimal

6.27358 \dots

Solution steps

\int_{1}^{3} \frac{6}{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}^{3} \frac{6}{x} d x - \int_{1}^{3} e^{- x} d x

\int_{1}^{3} \frac{6}{x} d x = 6 ln (3)

\int_{1}^{3} e^{- x} d x = \frac{1}{e} - \frac{1}{e ^{3}}

= 6 ln (3) - (\frac{1}{e} - \frac{1}{e ^{3}})

Simplify

= 6 ln (3) - \frac{1}{e} + \frac{1}{e ^{3}}

What is the integral from 1 to 3 of (6/x-e^{-x}) ?
The integral from 1 to 3 of (6/x-e^{-x}) is 6ln(3)-1/e+1/(e^3)