What is the integral from 1 to 2 of-1/x ?

The integral from 1 to 2 of-1/x is -ln(2)

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integral from 1 to 2 of-1/x

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

- ln (2)

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d 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_{0}^{\frac{π}{3}} e^{- t} cos (t) d t

\int_{1}^{2} x^{2} - \frac{1}{4 x ^{2}} d x

\int_{1}^{3} (\frac{3}{x ^{2}} - 1) d x

\int_{0}^{7 π} 5 + sin (x) d x

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

What is the integral from 1 to 2 of-1/x ?
The integral from 1 to 2 of-1/x is -ln(2)