What is the integral from b to a of 1/x ?

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

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integral from b to a of 1/x

\int_{b}^{a} \frac{1}{x} d x

ln ∣ a ∣ - ln ∣ b ∣

Solution steps

\int_{b}^{a} \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{b}^{a}

Compute the boundaries:

ln ∣ a ∣ - ln ∣ b ∣

= ln ∣ a ∣ - ln ∣ b ∣

\int_{0}^{l n (2)} 6 e^{3 x} d x

\int_{0}^{L} x^{2} d x

\int_{0}^{1} (5 x^{2} + x) d x

\int_{3}^{3.6} \frac{1}{x} d x

\int_{1}^{8} x^{2} x^{3} + 1 d x

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