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

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

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

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

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

Decimal

0.46060 \dots

Solution steps

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

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

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

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

Simplify

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

f (x) = e^{x}, at x = ln (11)

f (x) = 8 x - x^{2}

\frac{\partial}{\partial x} (ln (3 x^{2} + 2 y^{3}))

\int \frac{e ^{x}}{4 + 9 e ^{2 x}} d x

- x^{2} + 26 x + 920

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