What is the integral from y to 1 of 1/x ?

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

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integral from y to 1 of 1/x

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

- ln ∣ y ∣

Solution steps

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

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{y}^{1}

Compute the boundaries:

- ln ∣ y ∣

= - ln ∣ y ∣

\int_{0}^{π} (sin (x) - 2) d x

\int_{0}^{\frac{π}{2}} 6 + 3 cos (x) d x

\int_{0}^{6} \frac{1}{3} x e^{x} d x

\int_{2}^{- 2} (9 x + 8)^{2} d x

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

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