What is the integral from-e^6 to-e^3 of 1/x ?

The integral from-e^6 to-e^3 of 1/x is -3

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integral from-e^6 to-e^3 of 1/x

\int_{- e^{6}}^{- e^{3}} \frac{1}{x} d x

- 3

Solution steps

\int_{- e^{6}}^{- e^{3}} \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{- e^{6}}^{- e^{3}}

Compute the boundaries:

- 3

= - 3

\int_{2}^{3} (x^{3}) d x

\int_{0}^{k} x + 3 d x

\int_{2}^{\infty} v e^{- 3 v} d v

\int_{- 1}^{1} (1 + 6 x y^{2}) d y

\int_{0}^{4} - 6 x^{2} - 12 x + 144 d x

What is the integral from-e^6 to-e^3 of 1/x ?
The integral from-e^6 to-e^3 of 1/x is -3