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

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

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integral from x^1 to x^2 of 1/x

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

2 ln (x) - ln ∣ x ∣

Solution steps

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

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{x^{1}}^{x^{2}}

Simplify

= [ln ∣ x ∣]_{x}^{x^{2}}

Compute the boundaries:

2 ln (x) - ln ∣ x ∣

= 2 ln (x) - ln ∣ x ∣

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