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

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

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

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

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

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

Simplify

e^{- l n (x)} : \frac{1}{x}

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

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

\int_{4}^{13} \frac{1}{x - 4} d x

\int_{3}^{4} (x - 4)^{3} - (x - 4) d x

\int_{r^{1}}^{r^{2}} \frac{1}{r ^{2}} d r

\int_{0}^{1} \frac{1}{x ^{2} - x + 1} d x

\int_{0}^{2} \frac{x e ^{x}}{( 1 + x ) ^{2}} d x

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