What is the integral from-9 to-2 of 1/x ?

The integral from-9 to-2 of 1/x is ln(2)-2ln(3)

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integral from-9 to-2 of 1/x

\int_{- 9}^{- 2} \frac{1}{x} d x

ln (2) - 2 ln (3)

Solution steps

\int_{- 9}^{- 2} \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{- 9}^{- 2}

Compute the boundaries:

ln (2) - 2 ln (3)

= ln (2) - 2 ln (3)

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

\int_{x + 2}^{4 - x^{2}} 1 d y

\int_{0}^{2} (3 t^{3} + 5 t) d t

\int_{0}^{t} i (t) d t

\int_{1}^{9} x - 1 d x

What is the integral from-9 to-2 of 1/x ?
The integral from-9 to-2 of 1/x is ln(2)-2ln(3)