What is the integral from a to b of 1/x ?

The integral from a to b of 1/x is ln|b|-ln|a|

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integral from a to b of 1/x

\int_{a}^{b} \frac{1}{x} d x

ln ∣ b ∣ - ln ∣ a ∣

Solution steps

\int_{a}^{b} \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{a}^{b}

Compute the boundaries:

ln ∣ b ∣ - ln ∣ a ∣

= ln ∣ b ∣ - ln ∣ a ∣

\int - \frac{1}{x ^{4}} d x

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

\int 3 x^{2} - 2 x + 1 d x

d y + (2 x y - 4 e^{- x^{2}}) d x = 0

(4, 1), (3, 1)

What is the integral from a to b of 1/x ?
The integral from a to b of 1/x is ln|b|-ln|a|