What is the integral from-3 to-2 of 1/x ?

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

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integral from-3 to-2 of 1/x

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

ln (2) - ln (3)

Decimal

- 0.40546 \dots

Solution steps

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

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{- 3}^{- 2}

Compute the boundaries:

ln (2) - ln (3)

= ln (2) - ln (3)

\int 0.5 e^{- 0.5 x} d x

d er i v a t i v e \frac{c e ^{t}}{1 + c e ^{t}}

\frac{\partial}{\partial x} ((\frac{( x + 4 )}{y + 5})^{\frac{1}{2}})

\int (x^{2} + 7 x - 5) cos (2 x) d x

d er i v a t i v e f (x) = x (2 x - 1) (3 x + 2)

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