What is the integral from 17 to x of 1/t ?

The integral from 17 to x of 1/t is ln|x|-ln(17)

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integral from 17 to x of 1/t

\int_{17}^{x} \frac{1}{t} d t

ln ∣ x ∣ - ln (17)

Solution steps

\int_{17}^{x} \frac{1}{t} d t

Use the common integral:

\int \frac{1}{t} d t = ln (∣ t ∣)

= [ln ∣ t ∣]_{17}^{x}

Compute the boundaries:

ln ∣ x ∣ - ln (17)

= ln ∣ x ∣ - ln (17)

\int_{x}^{1} sin^{2} (t) d t

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\int_{0}^{2} t 1 + 4 t^{2} d t

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\int_{1}^{2} (x^{3} - 2) d x

What is the integral from 17 to x of 1/t ?
The integral from 17 to x of 1/t is ln|x|-ln(17)