What is the integral from 0 to x of 1/t ?

The integral from 0 to x of 1/t is diverges

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integral from 0 to x of 1/t

\int_{0}^{x} \frac{1}{t} d t

diverges

Solution steps

\int_{0}^{x} \frac{1}{t} d t

Use the common integral:

\int \frac{1}{t} d t = ln (∣ t ∣)

= [ln ∣ t ∣]_{0}^{x}

Compute the boundaries:

diverges

= diverges

\int_{0}^{\frac{8}{7}} \frac{1}{49 x ^{2} + 64} d x

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\int_{0}^{6} (4 t - 1)^{2} d t

\int_{3}^{6} (2 x - 5) d x

What is the integral from 0 to x of 1/t ?
The integral from 0 to x of 1/t is diverges